I found a solution for my problem. I post here a brief proof, in case anyone needs it.

I use the version of Alexander Duality, as stated in Bredon's book "Topology and Geometry" that states that if $M$ is a closed oriented $n$-manifold and $K\subset L$ are nice compact subspaces, then there is an isomorphism $$H_i(M\setminus{K}, M\setminus{L})\cong H^{n-i}(L, K)$$

**Step 1: If $F\subset D^4$ is a proper (not necessarily locally flat) surface in the $4$-disc, then $H_i(D^4\setminus{F}, S^3\setminus{\partial F})\cong H^{4-i}(D^4, F)$.**

Coning off $(S^3, \partial F)$ we obtain $S^4\supset X$, with $X=F\cup c(\partial F)$, where $c(\partial F)$ denotes the cone on $\partial F$. At this point we can use Alexander Duality to get
$$
H_i(S^4\setminus{X}, S^4\setminus{(D \cup X)})\cong H^{4-i}(D\cup X, X)
$$
where $D$ is the complement of a small open collar of $(S^3, \partial F)$ inside $D^4$.
To prove **Step 1** one simply notes that the inclusion
$$
(D^4\setminus F, S^3\setminus \partial F)\hookrightarrow (S^4\setminus{X}, S^4\setminus{(X\cup D)})
$$
induces isomorphisms in homology, and that by excision we have
$$
H^{4-i}(D\cup X, X)\cong H^{4-i}(D^4, F)
$$

**Step 2: If $L_1, L_2$ are two links in $S^3$ and $F\subset S^3\times I$ defines a (non necessarily locally flat) concordance between $L_1$ and $L_2$, then the inclusion of $\partial_-M=S^3\setminus L_1$ in $M=(S^3\times I)\setminus F$ induces a homology equivalence.**

Consider the cone on $(S^3, L_2)$ so to obtain $D^4\supset X$, where $X=F\cup c(\partial_+F)$. Notice that $\partial D^4\setminus \partial X=\partial_-M$.
Now apply step 1 to obtain
$$
H_i(D^4\setminus X, \partial_-M)\cong H^{4-i}(D^4,X).
$$
One obtains the thesis observing that $D^4\setminus X$ is homotopy equivalent to M, and that since both $D^4$ and $X$ are contractibles, the homology groups $H^{4-i}(D^4, X)$ all vanish.