Let $\zeta_1, \dots, \zeta_n$ be the roots of $p$ (not necessarily distinct). Let $R$ be big enough that all segments $[\zeta_i, \zeta_j]$ connecting distinct roots $\zeta_i \ne \zeta_j$ are contained in $U_R$.
I claim that $U_R$ is connected and simply connected.
1) $U_R$ is connected:
Suppose not. Then (by choice of $R$) all roots lie in one component of $U_R$. Let $V$ be a different component. We have $|p(z)| > 0$ for all $z \in V$ and $\overline V$ is bounded, hence compact. Therefore $p$ takes on a minimum in $\overline V$.
On the boundary $\partial V$ we have $|p(z)|\ge R$, so the minimum must lie in $V$. Since this minimum must be strictly bigger than $0$ by our choice of $V$, this is a contradiction to the maximum modulus theorem ($p$ is not constant).
This proves that $U_R$ must be connected.
2) $U_R$ is simply connected:
Again, suppose this was not the case. Then the complement has a bounded component $A$. Since $A$ is closed, $|p(z)|$ takes on a maximum in some point $z_0\in A$. Now, choose an open ball $B$ around $z_0$ which only intersects $A$ and possibly $U_R$ (such a ball exists because the complement $\mathbb C\setminus U_R$ of $U_R$ is closed and $A$ is disjoint from the rest of $\mathbb C\setminus U_R$).
But now $|p(z)|<R\le |p(z_0)|$ on $B\cap U_R$ and $|p(z)|\le |p(z_0)|$ on $B\cap A$ by choice of $z_0$.
This again contradicts the maximum modulus theorem, since $p$ is nonconstant.
This proves the claim.
The claim together with the Riemann mapping theorem shows that $U_R$ is homeomorphic to the unit disc.