No

Project the polyhedron in a planar graph. Suppose that the resulting graph has $V$ vertices, $F$ faces (counting the exterior face) and $E$ edges. We will prove by induction on $F$ that a polyhedron respecting the problem's conditions doesn't exist. For $F=2$ it's clear. Suppose that we have proved the statement for $F\le k$; we will prove it for $F=k+1.$ Case 1. There is an edge unused by both routes. In that case, suppose that such an edge borders faces $F_1,F_2$. Delete that edge and "merge" $F_1$ with $F_2$. The resulting planar graph has $k$ faces, and if $F_1$ and $F_2$ are initially imprinted with pairs $(x_1,y_1)$ and $(x_2,y_2)$, respectively, then the new face is imprinted with $(x_1+x_2,y_1+y_2)$. If $F$ satisfies the problem's condition, then so does this new graph, which would contradict the induction hypothesis. Case 2. Each edge is used by at least one ant. Let $W$ be the number of vertices of degree exactly $3$. Each route has at most $V$ vertices, hence at most $V$ edges. This implies that $2V\ge l_1+l_2$, where $l_i$ is the length of the route of ant $i$. Note that at any meeting point of exactly three edges, each route passes trough two of those edges, and so one edge belongs to both routes. Also, one such "repeated" edge is counted at most twice: at most once for each incident vertex. Putting all the pieces together, we obtain $$2V\ge l_1+l_2\ge E+\frac{W}{2}.$$ However, $$2E=\sum\text{vertex~ degrees}\ge 3W+4(V-W)$$i.e. $E+\frac{W}{2}\ge 2V.$ Thus, all inequalities are in fact equalities, so $l_1=l_2=V.$ If $(x_1=y_1),...,(x_{F-1}=y_{F-1}),(x_F\neq y_F)$ would be the pairs of numbers imprinted on the faces, then $$\sum_{i=1}^{F}x_i=2l_1=2l_2=\sum_{i=1}^{F}y_i$$i.e. $x_F=y_F$ as well, contradiction!

Let the two routes be $\ell_1$ and $\ell_2$. For every face $F$ of the polyhedron, denote $w_i(F)$ the the number of edges of $F$ belonging to $\ell_i$. We'll prove that it's impossible have exactly one $F$ with $w_1(F) \neq w_2(F)$. Clearly, a route $\ell_i$ divides the faces of polyhedron into two sides ("inside" and "outside"), say $A_i$ and $B_i$. For a face $F$, denote $$ u_i(F)= \begin{cases} 1 &\ (F \in A_i) \\ -1 &\ (F \in B_i) \end{cases}$$and let $u(F)=u_1(F)u_2(F)$. For an edge $E$ of the polyhedron, denote $$ v_i(E)= \begin{cases} 1 &\ (E \in \ell_i) \\ 0 &\ (E \notin \ell_i) \end{cases}$$and let $v(E)=v_1(E)-v_2(E)$. Thus for any face $F$, $$w_1(F)-w_2(F)=\sum_{E \subset F} (v_1(E)-v_2(E))=\sum_{E \subset F} v(E).$$Consider an edge $E$ satisfying that $v(E) \neq 0$, which means $E$ lies exactly one of $\ell_1$ and $\ell_2$. WLOG the former one, then if we let the two faces containing $E$ be $F$ and $G$, then $F$ and $G$ are in different sides of $\ell_1$, i.e. $u_1(F)u_1(G)=-1$. Similarly, $u_2(F)u_2(G)=1$. Thus $u(F)u(G)=u_1(F)u_1(G)u_2(F)u_2(G)=-1$, which also means $u(F)+u(G)=0$. In other word, for any edge $E$ we have $v(E)(u(F)+u(G))=0$. In conclusion, $$0=\sum_{E} v(E) \sum_{F: E \subset F} u(F) = \sum_{F} u(F) \sum_{E: E \subset F} v(E)=\sum_{F} u(F)(w_1(F)-w_2(F))$$Since $u(F) \in \{ -1,1 \}$, it's clearly impossible to have exactly one $F$ with $w_1(F) \neq w_2(F)$.

No

Project the polyhedron in a planar graph. Suppose that the resulting graph has $V$ vertices, $F$ faces (counting the exterior face) and $E$ edges. We will prove by induction on $F$ that a polyhedron respecting the problem's conditions doesn't exist. For $F=2$ it's clear. Suppose that we have proved the statement for $F\le k$; we will prove it for $F=k+1.$ Case 1. There is an edge unused by both routes. In that case, suppose that such an edge borders faces $F_1,F_2$. Delete that edge and "merge" $F_1$ with $F_2$. The resulting planar graph has $k$ faces, and if $F_1$ and $F_2$ are initially imprinted with pairs $(x_1,y_1)$ and $(x_2,y_2)$, respectively, then the new face is imprinted with $(x_1+x_2,y_1+y_2)$. If $F$ satisfies the problem's condition, then so does this new graph, which would contradict the induction hypothesis. Case 2. Each edge is used by at least one ant. Let $W$ be the number of vertices of degree exactly $3$. Each route has at most $V$ vertices, hence at most $V$ edges. This implies that $2V\ge l_1+l_2$, where $l_i$ is the length of the route of ant $i$. Note that at any meeting point of exactly three edges, each route passes trough two of those edges, and so one edge belongs to both routes. Also, one such "repeated" edge is counted at most twice: at most once for each incident vertex. Putting all the pieces together, we obtain $$2V\ge l_1+l_2\ge E+\frac{W}{2}.$$ However, $$2E=\sum\text{vertex~ degrees}\ge 3W+4(V-W)$$i.e. $E+\frac{W}{2}\ge 2V.$ Thus, all inequalities are in fact equalities, so $l_1=l_2=V.$ If $(x_1=y_1),...,(x_{F-1}=y_{F-1}),(x_F\neq y_F)$ would be the pairs of numbers imprinted on the faces, then $$\sum_{i=1}^{F}x_i=2l_1=2l_2=\sum_{i=1}^{F}y_i$$i.e. $x_F=y_F$ as well, contradiction!

No

Project the polyhedron in a planar graph. Suppose that the resulting graph has $V$ vertices, $F$ faces (counting the exterior face) and $E$ edges. We will prove by induction on $F$ that a polyhedron respecting the problem's conditions doesn't exist. For $F=2$ it's clear. Suppose that we have proved the statement for $F\le k$; we will prove it for $F=k+1.$ Case 1. There is an edge unused by both routes. In that case, suppose that such an edge borders faces $F_1,F_2$. Delete that edge and "merge" $F_1$ with $F_2$. The resulting planar graph has $k$ faces, and if $F_1$ and $F_2$ are initially imprinted with pairs $(x_1,y_1)$ and $(x_2,y_2)$, respectively, then the new face is imprinted with $(x_1+x_2,y_1+y_2)$. If $F$ satisfies the problem's condition, then so does this new graph, which would contradict the induction hypothesis. Case 2. Each edge is used by at least one ant. Let $W$ be the number of vertices of degree exactly $3$. Each route has at most $V$ vertices, hence at most $V$ edges. This implies that $2V\ge l_1+l_2$, where $l_i$ is the length of the route of ant $i$. Note that at any meeting point of exactly three edges, each route passes trough two of those edges, and so one edge belongs to both routes. Also, one such "repeated" edge is counted at most twice: at most once for each incident vertex. Putting all the pieces together, we obtain $$2V\ge l_1+l_2\ge E+\frac{W}{2}.$$ However, $$2E=\sum\text{vertex~ degrees}\ge 3W+4(V-W)$$i.e. $E+\frac{W}{2}\ge 2V.$ Thus, all inequalities are in fact equalities, so $l_1=l_2=V.$ If $(x_1=y_1),...,(x_{F-1}=y_{F-1}),(x_F\neq y_F)$ would be the pairs of numbers imprinted on the faces, then $$\sum_{i=1}^{F}x_i=2l_1=2l_2=\sum_{i=1}^{F}y_i$$i.e. $x_F=y_F$ as well, contradiction!