Suppose $f \in W^{1,1}_{loc}(U)$.  Then no, since for such an $f$, we have that $Df$ exists and the approximate limit

$ap\lim_{y\to x} \frac{f(x)-f(y)-Df(x)(x-y)}{|x-y|} = 0$

exists for almost every $x$, while from assuming classical differentiability we have

$\lim_{y\to x} \frac{f(x)-f(y)-\nabla f(x)(x-y)}{|x-y|} = 0$

exists for every $x \in U$.  In particular, the classical differential is a candidate for the approximate differential, and so $Df=\nabla f$ wherever the two exist, and hence in $U$ up to a set of measure zero.

http://www.encyclopediaofmath.org/index.php/Approximate_differentiability