At least, this is true if all roots of $p$ are pairwise distinct (and you do not need to assume anything about $q$ at all); this is why.
Writing $p(z)=C(z-z_1)\dotsb(z-z_n)$ with $z_1,\dotsc,z_n$ real shows that $p$ is a constant multiple of a polynomial with all its coefficients real. Normalizing, we can in fact assume that $C=1$, and so all coefficients of $p$ are real. Now choose a polynomial $s=\sum_{k=0}^n s_kz^k$ with all coefficients real, non-zero, and satisfying $s_kp_k\ge 0$. If, in addition, the coefficients $s_k$ are sufficiently small in absolute value, then $r(z):=p(z)+s(z)$ will have $n$ pairwise distinct real roots (as so has $p$), and the coefficients of $r$ will be between those of $p$ and of $q$ in the absolute value.
BTW, if all coefficients of $p$ are non-zero, then you can simply take $r=(1+\varepsilon)p$ with a sufficiently small $\varepsilon>0$. So, I suppose your question is essentially about the situation where some coefficients of $p$ vanish?