Suppose that $M$ is a compact, connected three-manifold with boundary. Suppose that $\pi_1(M)$ is infinite. Suppose that $\tilde{M}$, the universal cover of $M$, has at least three boundary components. Does $\tilde{M}$ have infinitely many boundary components?
Obviously the answer is "yes". I believe I have a proof of this, but my proof is annoyingly long and perhaps too machinery-heavy. It is also possible that I have missed some subcase of a subcase… Is there a clean (short, elementary, clever..) proof?
Non-counter-examples to keep in mind (for $M$) include handlebodies and interval bundles over closed surfaces. Another non-counter-example is $S^3$ minus a disjoint collection of (open, round) three-balls.
Best Answer
Let $M$ be a compact connected 3-manifold with nonempty boundary such that $\pi_1(M)$ is infinite and the universal cover $\tilde{M}$ has finitely many boundary components. I will prove that $\tilde{M}$ has either one or two boundary components.
Note: After writing this up I fear that this is just as tedious as what you describe (though it doesn't use any real technology), so it probably isn't so far off from what you had in mind.
Passing to a double cover, we can assume that $M$ is orientable.
Consider a component $B$ of $\partial M$. If the image of $\pi_1(B) \rightarrow \pi_1(M)$ has infinite index, then the preimage of $B$ in $\tilde{M}$ has infinitely many components. Since $\tilde{M}$ has finitely many boundary components, we conclude that the image of $\pi_1(B) \rightarrow \pi_1(M)$ is finite-index. Passing to a finite cover, we can assume that the map $\pi_1(B) \rightarrow \pi_1(M)$ is surjective.
Passing to a further finite cover might make $B$ separate into several components, but each of those components is still $\pi_1$-surjective. We can thus perform the procedure in the previous paragraph repeatedly and ensure that for each component $A$ of $\partial M$, the map $\pi_1(A) \rightarrow \pi_1(M)$ is surjective.
The proof now breaks up into two cases.
Case 1: $M$ has at least two boundary components.
Let $A$ and $B$ be two of the boundary components.
If $A$ is compressible, then we can find a compressing disk $D$ with $\partial D \subset A$. The disk $D$ must be separating; indeed, if it is nonseparating then we can find a loop $\gamma$ in $M$ that intersects $D$ once with positive sign. Since $\gamma$ can be homotoped into $B$, however, it can also be homotoped to be disjoint from $D$, a contradiction. Since $B$ lies on a single side of $D$ and $\pi_1(B) \rightarrow \pi_1(A)$ is surjective, we see that $\pi_1$ of the component of $M \setminus D$ not containing $B$ must be trivial. But this contradicts the fact that $\partial D$ is an essential loop in $A$ (reflect upon half-lives half-dies).
We deduce that $A$ is incompressible, so $\pi_1(A) \cong \pi_1(M)$. But it now follows from results in the chapter on I-bundles in Hatcher that $M$ is an $I$-bundle, so $\tilde{M}$ has two boundary components.
Case 2: $M$ has exactly one boundary component.
We are in the situation where $M$ is a compact oriented $3$-manifold with connected boundary such that the map $\pi_1(\partial M) \rightarrow \pi_1(M)$ is surjective. A standard exercise using the loop theorem shows that this means that $M$ is a handlebody, so $\tilde{M}$ has connected boundary.