Not always possible. Construction: $$A:=\begin{array}{|c|c|c|c|}\hline &&&D\\\hline &&&D\\\hline &D&&\\\hline &D&&\\\hline \end{array}$$where $D$s are dominoes. Cut the board into a $100\times100$ board on the left and a $100\times1$ board on the right. Use $25\times25$ $A$s to cover the $100\times100$ board. Color the board like this: $$\begin{array}{|c|c|c|c|c|c|c|c|c} \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\\\hline 1&1& 1& D& 0& 0& 0& D&\cdots\\\hline 1&1& 1& D& 0& 0& 0& D&\cdots\\\hline 1&D& 0& 0& 0& D&-1&-1&\cdots\\\hline 1&D& 0& 0& 0& D&-1&-1&\cdots\\\hline 0&0& 0& D&-1&-1&-1& D&\cdots\\\hline 0&0& 0& D&-1&-1&-1& D&\cdots\\\hline 0&D&-1&-1&-1& D&-2&-2&\cdots\\\hline 0&D&-1&-1&-1& D&-2&-2&\cdots\\\hline \end{array}$$We can see that when the token moving right or upward without passing a domino, the number in the cell can never decrease. The number in the bottom-left is $0$, but that in the top-right is $-1$, so it is impossible to move there.

Always possible. Let the bottom-left be $(1, 1)$ and the top-right be $(100, 100)$. Induction on $n$ to prove that if $(n, n)$ is free then it is always possible to go there from $(1, 1)$ without passing a domino. For $n=1$, it's trivial. Suppose for all $n\leq k$, it holds. For $n=k+1$, if $(k+1, k+1)$ is free, there are $2$ cases: Case 1: $(k, k)$ is free, since no $2$ dominoes are adjacent by vertices, at least one of $(k, k+1), (k+1, k)$ is free, and therefore the token can go to $(k, k)$ first by the induction hypothesis and then go to $(k+1, k+1)$. Case 2: $(k, k)$ is not free, WLOG suppose one of $(k, k-1), (k+1, k)$ is not free, since no $2$ dominoes are adjacent by vertices or edges, $(k-1, k-1), (k-1, k), (k-1, k+1), (k, k+1)$ are free, and therefore the token can go to $(k-1, k-1)$ first by the induction hypothesis and then go to $(k+1, k+1)$ through $(k-1, k), (k-1, k+1), (k, k+1)$. $\therefore$ by induction $(100, 100)$ is always possible to go from $(1, 1)$ without passing a domino.