Easy to see that $p(x)$ has even degree and positive leading coefficient. Then we can write $p(x) = u^2(x) + v(x)$, with $deg(v) < deg(u)$. Thus $p(p(x)) = (u(p(x)))^2 + v(p(x))$. But since $v(p)$ has lower degree than $u(p)$, this is a square only if $v(p) = 0$. Hence $v = 0$ and $p = u^2$.

Disclaimer: this solution is written late at night; hence it may contain errors; please notify me if so randomusername wrote: If $p,q\in\mathbb{R}[x]$ satisfy $p(p(x))=q(x)^2$, does it follow that $p(x)=r(x)^2$ for some $r\in\mathbb{R}[x]$? Claim: Let $A, B \in \mathbb{C}[X]$ be non-constant polynomials, then $\text{deg}(A^2-B^2)>0$ unless $A^2=B^2$. (Proof) Note that leading coefficient of $A$ cannot be equal to the leading coefficient of both $B, -B$; hence, one of $(A-B), (A+B)$ has positive degree. $\blacksquare$ WLOG, $\text{deg}(p)>0$ otherwise, the result is clear. Write $p=sr^2$ where $s, r \in \mathbb{R}[X]$ and $n=\text{deg}(s)$ is the minimal possible. For the sake of contradiction, let's suppose that $\text{deg}(s)>0$, then $$p(p)=q^2 \implies s(p)=t^2$$for some polynomial $t \in \mathbb{R}[X]$. Note that $p, s$ have positive leading coefficients. Note that $s$ is square free and we may write $$s(p(x))=\prod^n_{j=1} (p(x)-z_j),$$where $z_1, \dots, z_n$ are pairwise distinct complex roots of $s$. Note that $\gcd(p(x)-c, p(x)-d)=1$ over $\mathbb{C}[X]$ for $c, d \in \mathbb{C}$. Hence, there exists $t_1, \dots, t_n$ in $\mathbb{C}[X]$ so that $t=\prod^n_{j=1} t_j$ and for all $1 \le j \le n$, $$p(x)-z_j=t_j^2.$$If $n \ge 2$ then by our initial claim, $$t_1^2-t_2^2=-(z_1-z_2) \implies \text{deg}(t_1^2-t_2^2)=0,$$so, one of $t_1, t_2$ is constant (as $z_1 \ne z_2$), hence $p$ is constant, contradiction. However, if $n=1$ then $s(x)=(x-u)$ for some $u \in \mathbb{R}$, so $$t^2=s(p) \implies p(x)=t(x)^2+u \implies u=q(x)^2-t(p(x))^2.$$However, $$\text{deg}(q^2-t(p)^2)>0,$$again, a contradiction! To conclude, the answer is "yes", the claim above is necessarily true. $\blacksquare$

We work over $\mathbb C[x]$ for generality—our results carry smoothly down to $\mathbb R[x]$. Lemma: If $p_1^2-p_2^2\in \mathbb C$ for nonconstant polynomials $p_1,p_2$, then $p_1^2-p_2^2=0$. Proof: See Post #3. Suppose that $p\circ p=q^2$ for some $p,q\in \mathbb C[x]$. Write $p=\alpha \prod_{i=1}^k (x-r_i)^{m_i}$ with $\alpha\in \mathbb C$, the $m_i$ all positive integers, and the $r_i$ pairwise distinct elements of $\mathbb C$. Clearly this factorization is unique down to permutation of the indices $i$. Observe that for all $i\in\{1,2,\dots,k\}$, either $2\mid m_i$ or $p-r_i$ is a perfect square—or both. Also, $$|\{i:2\nmid m_i\}|= \sum_{i:2\nmid m_i} 1\equiv \sum_{i:2\nmid m_i}m_i\equiv \sum_{i=1}^k m_i=\text{deg}(q^2)\equiv 0\pmod{2} $$There are two cases: Case I: $|\{i\in \{1,2,\dots, k\}:2\nmid m_i\}|\geq 2$, which would imply the existence of $i_1\neq i_2$ such that $p-r_{i_1}$ and $p-r_{i_2}$ are both perfect squares, a contradiction to the Lemma. Case II: $|\{i\in \{1,2,\dots, k\}:2\nmid m_i\}|=0$, which would imply that $p$ is itself a perfect square. So the only possibility is that $p$ is a square $\blacksquare$