Here is a sketch of a proof:  Let $\dot f$ be a name for $f$.  Wlog we may assume that 
$\dot f$ is a nice name and hence uses only countably many conditions.
It follows that it is an $Fn(J,2)$-name for some countable set $J\subseteq I$.
Let $(p_n)_{n\in\omega}$ be an enumeration of $Fn(J,2)$.

We construct a function $g\in M$ as follows.
For each $n\in\omega$ choose $g(n)$ such that for all $k\leq n$ the following holds:
if for some $m\in\omega$ we have $p_k\Vdash\dot f(n)\gt m$, then $g(n)\gt m$.
It is possible to choose $g(n)$ in this way since for all $p\in Fn(J,2)$ there are only finitely many $m$ such that $p\Vdash\dot f(n)\gt m$.
 
We show that $g$ has the desired property.  Suppose this is not the case.
Let $p\in G$ and $n_0\in\omega$ be such that $$p\Vdash\forall n\geq n_0(\dot f(n)\gt g(n)).$$
For some $k\in\omega$, $p=p_k$.  Now let $n\geq n_0,k$.  Then $p_k\Vdash\dot f(n)\gt m$
for $m=g(n)$.  By the choice of $g$, $g(n)\gt m=g(n)$, a contradiction.