No. Suppose $m:\mathcal{P}([0, 1]) \to [0, 1]$ is a total extension of Lebesgue measure. Let $A \subseteq [0, 1]^2$ be such that every vertical section is Lebesgue null.

Claim: $A$ is $m \otimes \mu$-null.

Proof: For each rational $e>0$, let $U_{e, x} \subseteq [0, 1]$ be an open set of measure less than $e$ that contain the vertical section $A_x$. Let $G = \{(x, y) : (\forall e > 0)(y \in U_{e, x})\}$. For each rational interval $J$, let $X_{e, J} = \{x : J \subseteq U_{e, x}\}$. Then $G$ belongs to the sigma algebra generated by rectangles of the form $X_{e, J} \times J$ so $A$ is $m \otimes \mu$-null.

It follows that $\mu$-almost every horizontal section of $[0, 1]^2 \setminus A$ is not Lebesgue null. Note that this argument only requires the following: Every countably generated sigma algebra extending the Borel algebra admits a measure extending Lebesgue measure. A theorem of Carlson says that this holds in the random real model.