I suppose by $\mathbb{Z}_2$ you mean $\mathbb{Z}/2$ and not $\mathbb{Z}^{\wedge}_2$. Then we have $$H^*(BO(k);\mathbb{Z}/2)\cong
\mathbb{Z}/2[w_1,w_2\cdots w_k].$$ By definition we have
$\rho ^*(w_n)=w_n(r)$ where $w_n(r)$ is the $n$-th Stiefel-Whiteney class of the
representation $r$. Since the regular representation factors through $O(k-1)$, we see that $w_k=0$. We can, in theory, determine the image using the fact that
the Quillen's map is injective (as a matter of fact, isomorphic) for the symmetric groups, that is elements of mod $2$ cohomology of the symmetric get detected by its elementary abelian subgroups.

In the case of $k=4$, for example,
if we call $V_1=\{(12),(34),(12)(34),id\},V_2=\{(12)(34),(13)(24),(14)(23),id\}$
then the restriction of $w_n(r)$'s to $BV_1$ can be computed using Whitney product formula, and the restriction to $BV_2$ can be found more or less using the definition of Dickson invariants, so by comaparing with the table in http://homepages.math.uic.edu/~bshipley/ConMcohomology1.pdf Example 4.4, we see that they coincide with standard generators of $H^*(B\Sigma _4;\mathbb{Z}/2)$. Thus for $k=4$ $\rho ^*$ is surjective. In principle, this kind of analysis can be carried out for larger $k$, but I am not aware of any explicit result.