Yes, (and of course): The very definition of the étale fundamental group is that it classifies finite étale covers.
Precisely: Let $X$ be a connected scheme and $x$ be a geometric point of $X$. There is by construction an equivalence of categories between finite $\pi_1^{\mathrm{ét}}(X,x)$--sets and finite étale coverings of $X$, connected coverings corresponding to transitive sets. This property characterises $\pi_1^{\mathrm{ét}}(X,x)$ up to unique isomorphism.
A universal cover as in topology also exists, although only as a profinite cover.
A fine source for this is T. Szamuely's book "Fundamental groups and Galois groups" (Cambridge 2009).
Throughout the following, I'll say "homotopy category" when I really mean the weak homotopy category.
For a space $X$, the homology of $X$ is canonically isomorphic to the reduced homology of $X_+$, which is $X$ with a disjoint basepoint. Therefore, it suffices to give a definition of the reduced homology of a based space.
The smash product $\wedge$ descends to a well-defined operation on the homotopy category of based topological spaces. For any $n \geq 0$, we have an object $K(\mathbb{Z},n)$ in the homotopy category, and weak equivalences $K(\mathbb{Z},n) \to \Omega K(\mathbb{Z},n+1)$ which are adjoint to maps $S^1 \wedge K(\mathbb{Z},n) \to K(\mathbb{Z},n+1)$. For any integer $m$, we can therefore define a directed system of sets
$$
[S^{m+k}, K(\mathbb{Z},k) \wedge X ] \to
[S^{m+k + 1}, K(\mathbb{Z},k+1) \wedge X ] \to \cdots
$$
The colimit $colim_k \pi_{m+k}( K(\mathbb{Z},k) \wedge X)$ is isomorphic to the $m$'th reduced homology group of $X$ in a canonical way.
This kind of definition produces generalized homology theories, and this all falls into the subject of stable homotopy theory.
The method you suggest of taking the free spectrum in this $\infty$-categorical sense will give a homology theory, but instead of producing the homology groups of $X$ it will produce the stable homotopy groups. The "free abelian topological group" on the topological space $X$ can be used instead, and shown to give a good notion on the homotopy category; this will produce the homology of $X$, as a result of the work of Dold and Thom.
Best Answer
This is certainly false! One should look at the corresponding group theory problem: any group $G$ is the fundamental group of a manifold $M$, and the first homology of $M$ is $\pi_1(M)^{ab} = G^{ab}$. Thus the problem becomes: given a group $G$ and a finite index subgroup $H$ of index $p$, does $G^{ab}$ torsion free imply that $H^{ab}$ is torsion free ( edit $p$-torsion free)? (It is unclear whether you are insisting that the $p$-covers be Galois, which would correspond to insisting that $H$ is normal, but both variations have negative answers.)
For an explicit example, let $G = \langle x,y \ | \ [x,y]^2 \rangle$. Then $G^{ab} = \mathbf{Z}^2$, but the homomorphism $G \rightarrow \mathbf{Z}/2$ sending $x$ and $y$ to $1$ has kernel $H = \langle a,b,c \ | \ (cb^{-1}a^{-1})^2, (c^{-1}ba)^2 \rangle$, where $a = yx^{-1}$, $b = x^2$, and $c = xy$. In particular, we see that $H^{ab} = \mathbf{Z}^2 \oplus \mathbf{Z}/2$.
A previous version discussed knot complements and the Alexander polynomial, but I had missed read "torsion free" for "$p$-torsion free", and so the example does not apply.
One positive remark in the direction of your question: If you are assuming that $H$ is normal, then $H^{ab}$ is $p$-torsion free if the rank of $G^{ab}$ is less than two. This is because the latter condition implies that the $p$-completion of $G$ is cyclic, a condition which is inherited by a normal subgroup of index $p$.
The methods used to compute homology vary considerably depending on what information you have. If you have a presentation for the group, you can use the Reidemeister--Schreier algorithm to compute a presentation of $H$, from which it is easy to compute $H^{ab}$. The more you understand the geometry of the situation, however, the better.