The answer is $k=4$. This can be obtained with the following construction: [grid image]. It now suffices to show that $k=5$ and $k=6$ are not attainable. The case $k=6$ is clear. Assume for sake of contradiction that the $k=5$ is attainable. Let $r_{1}, r_{2}, r_{3}$ be the rows of three distinct uncolored cells, and let $c_{1}, c_{2}, c_{3}$ be the columns of the other three uncolored cells. Then we can choose $R_{1}, R_{2}, R_{3}$ from $\{1,2,3,4,5,6\} \backslash\left\{r_{1}, r_{2}, r_{3}\right\}$ and $C_{1}, C_{2}, C_{3}$ from $\{1,2,3,4,5,6\} \backslash\left\{c_{1}, c_{2}, c_{3}\right\}$ to obtain a contradiction.