If $q\colon\mathbb{C}\to\mathbb{C}$ is a polynomial, $f\colon\mathbb{C}\to\mathbb{C}$ is analytic on all of $\mathbb{C}$, and if there exists $a\gt 0$ such that $|f(z)| \lt a|q(z)|$ for every $z\in \mathbb{C}$, then $f = bq$ for some $b\in \mathbb{C}$.
Can an arbitrary analytic function (on all of $\mathbb{C}$) replace $q$?
If you really mean strict inequality, then this follows from Liouville's theorem applied to $f/q$. Note that $q$ must be constant too, as $q$ can have no zeros from the condition.
It is actually true if $<$ is replaced by $\leq$, though. (Maybe you meant this?) First, the Cauchy estimates show that $f$ is a polynomial; indeed, since $f$ grows polynomially, we can just take averages of $f/(z-\alpha)^N$ over larger and larger circles. From this one can see that $f$ is a polynomial.
Now the question reduces to showing that if a polynomial $p$ is bounded by a constant multiple of another polynomial $q$, then $p$ and $q$ differ by a constant. This is a straightforward consequence of factorization.
