As mentioned by David, the answer is "no": If $\Sigma$ is the Riemann sphere and $\mathfrak{p}$ consists of a single point, then there is no such covering. (Indeed, in this case $\Sigma\setminus\mathfrak{p}$ is simply-connected, so there is only one ramified cover, and this cover has degree one ...

If you have more than one point, or $\Sigma$ is not the sphere, then the answer is "yes". In the case of the sphere with two points, you can use $z\mapsto z^d$ (up to conformal change of coordinate), and this is the only choice.

Otherwise, your cover $S$ can be chosen to be the unit disc. Indeed, consider the surface as an "orbifold" with given some ramification indices at your given points. It is known that this orbifold has a universal covering, which means that there is a holomorphic function, on the sphere, plane or disc, which is ramified in each preimage of the ramification points, with multiplicity a multiple of the given ramification index. In most cases, in particular if the ramification indices are taken sufficiently large, the universal cover will be the unit disc, as claimed.

This argument also works for infinite but discrete subsets of $\Sigma$.

In fact, it is not hard to see that the result will be true for any countable collection of branched values. (Here of course the map will not be an orbifold covering. However, using the orbifold covering argument we first see that we prove the result for the disc and a discrete set of points, and then apply the orbifold argument again ...)

(Of course, even if the surface $\Sigma$ is compact, this argument gives a non-compact cover - which may or may not be what you had in mind.)