There are $27$ multiples of $3$ from $1$ through $81$. A row or column has product divisible by $3$ exactly when it contains at least one of these multiples.

Suppose $r$ rows and $c$ columns have products divisible by $3$. Every multiple of $3$ must lie in one of those $r$ rows and also in one of those $c$ columns, or else it would create another marked row or column. Thus the $27$ multiples must fit in the $rc$ intersection cells. If $r+c\le10$, then $rc\le25$, which is too small. So at least $11$ rows and columns are needed.

This can be done by placing $25$ multiples of $3$ in a $5\times5$ block, then placing the remaining $2$ multiples in a sixth column within two of those same rows. Then exactly $5$ rows and $6$ columns are marked, for a total of $11$.

Thus, D is the correct answer.