No. 


Let $K$ be a Cantor set in the unit interval of positive one-dimensional Lebesgue measure. More generally, we can take any measurable subset of $\bf R$ of positive Lebesgue measure and empty interior. Rotate $K\times K \subset {\bf R}^2$ by a quarter turn (45 degree) in the plane.  The resulting set cannot contain a subset $A\times B$ with $A$ and $B$ of positive measure.

This is shown as follows. Let us project our set on the line of slope -1 through the origin, graduated so that the point $(x,y)$ is sent to the point $x-y$ on the line.

It is a standard fact that if $A$ and $B$ are two subsets in $\bf R$ of positive Lebesgue measure, then the set $A-B=\{x-y \mid x\in A, \ y\in B\}$ contains an interval. This follows from the continuity of $x\mapsto \int {\bf 1}_A(t-x) {\bf 1}_B(t) \ dt$.

So the projection of our set on the line must contain an interval. But this projection is (a translation-rotation of) $K$ which is of empty interior.