Yes, this can happen.  A little experimentation with the structure equations shows that there is a $3$-parameter family of homogeneous examples in dimension $4$:  Let $c_1,\ldots,c_4$ be constants satisfying $c_1c_2=c_3c_4\not=0$, and consider the simply-connected $4$-dimensional Lie group $G$ that has a basis for the left-invariant $1$-forms given by $\omega_1,\ldots,\omega_4$, which satisfy the structure equations
\begin{aligned}
d\omega_1 &= 2c_1\ \omega_2\wedge\omega_3 + 2c_3\ \omega_4\wedge\omega_3\ ,\\\\
d\omega_2 &= 2c_1\ \omega_3\wedge\omega_1 \ ,\\\\
d\omega_3 &= 2c_2\ \omega_4\wedge\omega_1 + 2c_4\ \omega_2\wedge\omega_1\ ,\\\\
d\omega_4 &= 2c_2\ \omega_1\wedge\omega_3 \ .
\end{aligned}

Now endow $G$ with the Riemannian metric for which the $\omega_i$ are an orthonormal basis and let $\mathcal{E}$ be the $2$-plane field defined by $\omega_1=\omega_2=0$ while $\mathcal{F}$ is the $2$-plane field defined by $\omega_3=\omega_4=0$.  One easily checks that this is an example of the desired type.  For example, if $\nabla$ is the Levi-Civita connection of this metric and $e_1,\ldots, e_4$ is the dual basis of left-invariant vector fields, then one has (all congruences modulo $\{e_1,e_2\} = \mathcal{F}$ )
$$
\nabla_{e_1}e_1\equiv0\ ,\quad\nabla_{e_1}e_2\equiv c_4e_3, \quad
\nabla_{e_2}e_1\equiv-c_4e_3\ ,\quad\nabla_{e_2}e_2\equiv 0.
$$

There are non-homgeneous examples in dimension $4$ as well.  A little more work with the structure equations shows that there exists a $4$-parameter family of examples of cohomogeneity $2$.  (I don't know how many of these are complete.)  If there is interest, I can give the structure equations of these examples as well.

It is not clear that there are any other examples in dimension $4$.