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 nonzero constants satisfying $c_1c_2=c_3c_4$, and consider the simply-connected $4$-dimensional Lie group $G$ that has a basis of left-invariant $1$-forms $\omega_1,\ldots,\omega_4$ that 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 $g$ for which the $\omega_i$ define an orthonormal coframing, let $e_1,\ldots,e_4$ be the dual ($g$-orthonormal) vector fields, and let $\mathcal{E}$ be the $2$-plane field spanned by $e_1$ and $e_2$ while $\mathcal{F}$ is the $2$-plane field defined by $e_3$ and $e_4$.

One easily checks that this is an example of the desired type: If $\nabla$ is the Levi-Civita connection of this metric, then
$$
\nabla_{e_1}e_1\equiv0\ ,\quad\nabla_{e_1}e_2\equiv c_4e_3, \quad
\nabla_{e_2}e_1\equiv-c_4e_3\ ,\quad\nabla_{e_2}e_2\equiv 0 \mod \mathcal{E}
$$
and
$$
\nabla_{e_3}e_3\equiv0\ ,\quad\nabla_{e_3}e_4\equiv c_3e_1, \quad
\nabla_{e_4}e_3\equiv-c_3e_1\ ,\quad\nabla_{e_4}e_4\equiv 0 \mod \mathcal{F},
$$
as desired.

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.