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.
Best Answer
In general working with infinite collections of maps is harder. As far as I know, the main result in this direction is the "disc deployment lemma" of Frank Quinn. Roughly, it says that for any $\epsilon>0$ there exists a $\delta>0$ such that an infinite collection of immersed discs, with dual spheres, controlled by $\delta$, can be replaced by an embedded collection, controlled by $\epsilon$. This is a very powerful result, which Quinn used to prove fundamental results about 4-manifolds, such as topological transversality and that noncompact 4-manifolds are smoothable.
The original source for this result is Frank Quinn's paper titled "Ends of Maps III: dimensions 4 and 5". published in the Journal of Differential Geometry. It is a challenging paper to read. You might also consider looking at the treatment by Edwards in "The solution of the 4-dimensional annulus conjecture (after Frank Quinn)", published in the conference proceedings of a conference called "Four-manifold theory (Durham, N.H., 1982) (the official citation is: Four-manifold theory (Durham, N.H., 1982), 211–264, Contemp. Math., 35, Amer. Math. Soc., Providence, RI, 1984). In the book by Freedman and Quinn, the relevant theorem is on page 90.