What you can prove is that if you take a $2$-complex $X$ with fundamental group $A\ast B$ and you take a finite subcomplex $K$ of a covering $X'$ of $X$ such that the fundamental group of $K$ maps under the covering to a subgroup of $A\ast B$ which is closed in the profinite topology, then the restriction of the covering map to $K$ can be extended to a finite sheeted covering. Moreover, you can do it so that the Kurosh decomposition of your finitely generated closed subgroup is compatible with the Kurosh decomposition of the finite index subgroup (e.g., the conjugates of factors are contained in conjugates of factors and the free part of the smaller group is a free factor of the free part of the larger group).
A topological proof in the case $A,B$ are subgroup separable (but the proof can be made to work under the weaker assumption above) is given in our paper http://www.sciencedirect.com/science/article/pii/S0022404902002104
An equivalent formulation in the language of Bass-Serre theory can be found in the paper of Ribes and Zalesskii PROFINITE TOPOLOGIES IN FREE PRODUCTS OF GROUPS