Yes, there does. In fact, we will show that a polynomial $q(x) = x^2 + d$ works for some $d \in \mathbb{Q}$. The conclusion will follow from Hilbert's Irreducibility Theorem once we show that $p(x^2 + d)$ is irreducible (as a polynomial in two variables).

Suppose, therefore, that $p(x^2 + d) = (d^a f_a(x) + \dotsb)(d^bg_b(x) + \dotsb)$ splits. Both $a$ and $b$ are positive (else you could plug in some value of $x$ in $\mathbb{C}$ to make the RHS vanish regardless of $d$, which is impossible by starring at the LHS). But then $f_a$ and $g_b$ are nonzero and so there is a rational $x$ which is a zero of neither of them. Plug it in and get a nontrivial splitting of (a shift of) $p(x)$. Contradiction, as $p(x)$ was assumed to be irreducible.