The short answer to your question is that the Cantor set corresponds to the limit set of the action of $\pi_1 S$ on $\mathbb{H}^2$.
Here are more details, the idea being that the topological description of the universal cover $\tilde S$ is a consequence of a more detailed geometric description which can be found in any basic textbook on hyperbolic manifolds, such as the book of Thurston et.al. or the book of Ratcliffe.
The first basic fact one needs is that the universal cover $\tilde S$ of $S$ with its lifted hyperbolic structure embeds isometrically in $\mathbb{H}^2$, so that the deck transformation action of $\pi_1(S)$ on $\tilde S$ extends to a free and properly discontinuous action of $\pi_1(S)$ on all of $\mathbb{H}^2$ by isometries.
Next, one needs the concept of the limit set $\Lambda$ of the action of $\pi_1(S)$ on $\mathbb{H}^2$, which in this setting is equal to the set of points in the circle at infinity $S^1_\infty = \partial \mathbb{H}^2$ which are contained in the closure of $\widetilde S$ relative to the closed disc $\overline{\mathbb{H}}^2 = \mathbb{H}^2 \cup S^1_\infty$. This closure of $\tilde S$ is equal to the union $\tilde S \cup \Lambda$, and is also equal to the convex hull of the set $\Lambda$ which means the smallest closed subset of $\overline{\mathbb{H}}^2$ which contains $\Lambda$ and contains the hyperbolic geodesic segment between any two of its points. It follows from this description that $\tilde S \cup \Lambda$ is homeomorphic to the closed 2-disc, and it also follows that $\Lambda$ is on the boundary of that 2-disc.
What remains, then, is to see that $\Lambda$ is homeomorphic to the Cantor set. One shows directly, using the fact that $\Lambda$ is the limit set of the action $\pi_1(S)$, that $\Lambda$ is perfect and totally disconnected. Then one applies the basic theorem that every perfect, totally disconnected, nonempty, compact metric space is homeomorphic to the Cantor set.
Let me also correct a few mis-impressions in your question.
As said in my answer above, the universal cover $\tilde S$ is embedded in $\mathbb{H}^2$ which is the interior of the Poincare disc. But $\tilde S$ is a proper subset of the interior of the Poincare disc, so $\tilde S$ is neither equal to the whole of the Poincare disc (because $\tilde S$ is disjoint from the boundary of the Poincare disc), nor is it equal to the whole of the interior of the Poincare disc.
Also, In this process you should not cap off the boundary components of $S$. Each component of $\partial S$ is a geodesic circle, and each component of $\partial \tilde S$ is a geodesic line in $\mathbb{H}^2$. There is a countably infinite number of components of $\partial \tilde S$. The universal covering map $\tilde S \to S$ restricts to a universal covering map from each component of $\partial \tilde S$ to some component of $\partial S$.
