Probably I'm blind now, but what is wrong with the following argument: for any $z$ on the unit circle we have $P(z)Q(\frac{1}{z})=R(z)$. This is an identity between rational functions, because it holds for infinitely many $z$. Similarly, by taking $z$ on the circle of radius $2$ we find that $P(z)Q(\frac{4}{z})=R(z)$ for all $z$. Thus because $P$ is nonzero, we deduce that $Q(z)=Q(4z)$ and thus $Q$ is constant. What is wrong with this?

The statement is true in the general case, when $P,Q,R$ are complex polynomials. In the plane, consider the differential operators $\frac\partial{\partial z}=\frac{1}{2}\left(\frac\partial{\partial x}-i\frac\partial{\partial y}\right)$ and $\frac\partial{\partial \bar z}=\frac{1}{2}\left(\frac\partial{\partial x}+i\frac\partial{\partial y}\right)$. The condition for a function $f$ to be holomorphic is $\frac{\partial f}{\partial\bar z}=0$. Since $Q(\bar z)=\frac{R(z)}{P(z)},\ z\mapsto f(z)=Q(\bar z)$ is holomorphic in any domain noit containing the roots of $P$. The map $z\mapsto\bar f(z)$ is also holomorphic, because it's a polynomial in $z$ having as coefficients the conjugates of the coefficients of $Q$. This means that on the one hand we have $\frac{\partial f}{\partial\bar z}=0$, and on the other hand we have $\frac{\partial\bar f}{\partial\bar z}=0\Rightarrow\frac{\partial f}{\partial z}=0$. We thus get $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial z}+\frac{\partial f}{\partial\bar z}=0$, and similarly, $\frac{\partial f}{\partial y}=0$, which implies that $f$ is constant. Edit: I was a bit too late I guess .

Take $j=0,1,2,...$ successively in the statement $\forall r \in \mathbb{R}, Q(i^{\frac{1}{2^j}}r)=Q((-i)^{\frac{1}{2^j}}r)$ to prove the result for $Q\in \mathbb{C}[x]$ as this means that the coefficients of $x^k$ in $Q$ where $\nu_2(k) = 0,1,2,...$ are 0.

We claim that $\overline{Q}\in\mathbb C[z]$. Suppose (in the interest of contradiction) that $P\nmid R$, and note by the Fundamental Theorem of Algebra that there exists some $\zeta\in\mathbb C$ for which $P(\zeta)=0\neq R(\zeta)$. It follows that $P(\zeta)Q(\overline{\zeta})=0\neq R(\zeta)$, impossible. Now observe that $Q\cdot \overline{Q}$ takes values only in $\mathbb R$. It is well-known (and easily shown) that any such polynomial is constant, and thus so too must $Q$ be $\blacksquare$