The answer to your first question is 'yes, generically, the intersection is finite and transverse'. More precisely, the set $U$ of 6-dimensional subspaces that intersect $\mathrm{SO}(3)$ transversely (and hence in a finite number of points), is a dense open set in $\mathrm{Gr}(6,9)$. (However, $U$ is not connected; the number of intersection points can be (and are) different for different components of $U$.) For example, the $6$-dimensional subspace $V_+$ consisting of upper triangular matrices intersects $\mathrm{SO}(3)$ transversely in the $4$ points that consist of the diagonal elements of $\mathrm{SO}(3)$. Thus every $6$-dimensional subspace sufficiently near $V_+$ also intersects $\mathrm{SO}(3)$ in $4$ points and transversely.
As for your parenthetical question, the answer to that is 'no, there exist $6$-dimensional subspaces $W$ that do not meet $\mathrm{SO}(3)$ and for which the generic element in $W$ is invertible'. For example, the generic $W$ that is sufficiently near the $W_0$ consisting of the matrices with zero first column will have its generic element be invertible.