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 $p_k\Vdash\dot f(n)=m$ for some $m\in\omega$, then $g(n)\geq m$.

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.