Zarrax has already given a great example. Here's a way to see that there are polynomial examples from the fact that there are holomorphic examples. The motivation for this is that the non-polynomial example $g(z)=e^{2\pi i z}$ came to mind when I read your question.
Let $g$ be holomorphic in a neighborhood of the closed disk, not injective on the open disk, but with nonvanishing derivative on the closed disk. Let $(p_n)_n$ be the sequence of partial sums of the Taylor series of $g$ centered at $0$, so $p_n\to g$ and p_n'\to g' uniformly on a (smaller) neighborhood of the closed disk.
Since g' is nonvanishing on the closed disk, it has a positive minimum modulus $m$ there, which by uniform convergence means that p_n' eventually has modulus greater than $m/2$, and in particular is eventually nonvanishing on the closed disk. Take $a\neq b$ in the open unit disk such that $g(a)=g(b)$. Applying Hurwitz's theorem to the sequence of functions $p_n(z)-g(a)$ on small disjoint disks centered at $a$ and $b$ shows that $p_n$ eventually takes on the value $g(a)$ at more than one point in the open unit disk. Thus, $p_n$ is eventually a counterexample.
In fact, using WolframAlpha, it looks like $f(z)=\displaystyle{\sum_{k=0}^{25}\frac{(2\pi iz)^k}{k!}}$ gives an example. It allegedly takes on the value $-1$ at points very close to $\pm\frac{1}{2}$, and all $24$ of the zeros of its derivative are outside the closed disk.