Faster growing busy beaver function
Suppose there is a machine using $n$ states which, at some point, has $X$ nonzero symbols on the tape. One can build a machine with $O(n)$ states which simulates a run of the original machine with a tape alphabet of three symbols $\{0,1,2\}$, with the following small change: whenever the original machine changes $1$ to $0$, the new machine changes it to $2$; whenever the tape is read, $2$ is interpreted as $0$. The new machine ends up with $Y \in [X,2X]$ nonzero symbols on the tape (we get $\Theta(X)$ instead of $X$ since the simulation requires using two symbols for each original symbol). So your new function, let's call it $F(n)$, satisfies $B(n) \leq F(n) \leq B(O(n))$.