(Solution by A. Asanau, D. Voynov.) Replace $x$ by $1-x$ in
$$p(x^2) = p(x)q(1-x) + p(1-x)q(x), \quad (1)$$
then we have $p((x-1)^2) = p(1-x)q(x) + p(x)q(1-x) = p(x^2)$. It follows that the polynomial $p(x^2)$ is periodic, i.e., $p(x)$ is a constant polynomial, $p(x) = a$ for some $a \in \mathbb{R}$.
If $a=0$, then (1) is valid for any arbitrary polynomial $q(x)$.
If $a \neq 0$, then (1) becomes $q(1-x) + q(x) = 1$ for all $x$. Replacing $x$ by $x+0.5$ we can rewrite the last equality as
$$q(0.5 - x) - 0.5 + q(0.5 + x) - 0.5 = 0. \quad (2)$$
Let $h(x) = q(0.5 + x) - 0.5$. Then (2) shows that $h(-x) = -h(x)$, so we may write $h(x) = xr(x^2)$ for some polynomial $r(x)$.
Hence $q(0.5 + x) = xr(x^2) + 0.5$. Replace $x$ by $x - 0.5$ here, we obtain $q(x) = (x - 0.5)r((x - 0.5)^2) + 0.5$ which obviously satisfies the condition.