Let us call a set $A\subset \mathbb{R}^2$ uniformly grey if the measure of its sections is constant, but not full. (There may be a standard name for this; I would be glad if someone tells me.)
Formal definition: There are intervals $[a_1,b_1]$ and $[a_2,b_2]$ and constants $\mu_1\in(0,\lambda(A_2))$ and $\mu_2\in(0,\lambda(A_1))$ such that $\lambda(\{x_2:(x_1,x_2)\in A\}) = \mu_1$ for any $x_1\in (\inf a_1,b_1)$ and $\lambda(\{x_1:(x_1,x_2)\in A\}) = \mu_2$ for any $x_2\in (a_2, b_2)$; $\lambda$ is the Lebesgue measure.
A simple example is $A = \{(x_1,x_2)\in[0,1]^2: (x_2-x_1)\mod 1\in(0,1/2)\}$.
Well, this does not look grey at all, so in order to make it look grey, let us add an assumption of positive density: for any $x = (x_1,x_2)\in (a_1,b_1)\times (a_2,b_2)$ and any $r>0$, $\lambda_2(B(x,r)\cap A)>0$, where $B(x,r)$ is the ball of radius $r$ centered at $x$.
So the question is:
Does there exist a uniformly grey set of positive density?
Take a fat Cantor set $C$ (such as the Smith-Volterra-Cantor set) in $[0,1]$ of measure $\frac{1}{2}$. Then $C \times C$ will satisfy your section condition.
Does it have positive density? The points in the fat cantor set have positive density on $\mathbb{R}$, and so given any ball around our point we can find a square within that ball having positive intersection with our set.