If you have an oracle coding not only the atomic truth of the model but also the $\Delta_0$ diagram, then yes, you can compute an oracle for the forcing extension: given any poset $P$ in that model, we can compute a model for a for forcing extension $M[G]$, with $M$-generic $G\subset P$. The reason is that the oracle for $M$ includes a way for us computably to enumerate all the dense subsets of $P$ in $M$, and from that, we may computably construct an $M$-generic filter $G\subset P$, by building a descending $\omega$-sequence and selecting the first element we find in the next dense set. Given the oracle for $M$ and our computation of $G$, we can now compute equality of names relative to this filter, precisely because we can compute $\Delta_0$ truth in the model, and we can thereby compute a representation of $M[G]$.
In the end, the Turing degree of the original representation of $M$ and the representation of $M[G]$ will be the same; they are computable from each other.
I've got to think a bit more about the case where you only have $\in$ and not the $\Delta_0$ diagram. You can still compute a filter $G$ in this case, but I don't see how to compute equivalence of names without access to more than just $\in^M$.