If you have an oracle coding not only the atomic truth of the model but also the $\Delta_0$ diagram, then yes, given any poset $P$ in that model, we can compute a representation of a forcing extension $M[G]$, with $M$-generic $G\subset P$. The main 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 $G\subset P$. Specifically, $P$ is represented in the oracle by some number $P$; the relation $\leq_P$ is also represented by some number, and the collection $D$ of all dense subsets of $P$ in $M$ is represented by some number. Now, we computably construct a sequence $p_n$ which will be descending in $P$ and meet every dense set in $M$, simply by searching for the next element of the oracle that is in $P$ and in the next thing we find that is in $D$, and such that it is $\leq_P$ related to $p_n$ (this requires us to find the object in $M$ coding the pair, etc.). We don't need the $\Delta_0$ diagram to compute $G$. But now, having computed an $M$-generic filter $G$, we can compute an oracle for $M[G]$ as follows: we can use the $\Delta_0$ oracle to tell when an element of $M$ is a $P$-name and when two names are forced to be equal by a condition in $G$ (the particular conditions we put into $G$ will either force them to be equal or force them to be unequal). So we can build a computable representation of $M[G]$ by using the names as indices, and using our decision procedure for equality modulo $G$ to avoid double representation, and then using the fact that $p\Vdash \tau\in\sigma$ is $\Delta_0$ to compute the relation $\in^{M[G]}$ for our representation.

Similarly, if we have an oracle for the full elementary diagram of $M$, then we can compute an oracle for the full elementary diagram 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.

In the case that the oracle only gives you access to $\in^M$, however, and not the $\Delta_0$ diagram of $M$, then although you can still compute a particular generic filter $G$, it isn't clear to me that we should expect to be able to compute equality of names modulo $G$, and so I suspect that one might not be able to compute a representation of $M[G]$ in this general case.