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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 $c_1c_2=c_3c_4\not=0$, c_1c_2=c_3c_4$, and consider the simply-connected $4$-dimensional Lie group $G$ that has a basis for the of left-invariant $1$-forms given by $\omega_1,\ldots,\omega_4$, which \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 $\omega_i$ are define an orthonormal basis coframing, let $e_1,\ldots,e_4$ be the dual ($g$-orthonormal) vector fields, and let $\mathcal{E}$ be the $2$-plane field defined spanned by $\omega_1=\omega_2=0$ e_1$ and $e_2$ while $\mathcal{F}$ is the $2$-plane field defined by $\omega_3=\omega_4=0$. e_3$ and $e_4$.

One easily checks that this is an example of the desired type. For example, if : If $\nabla$ is the Levi-Civita connection of this metricand $e_1,\ldots, e_4$ is the dual basis of left-invariant vector fields, then one has (all congruences modulo $\{e_1,e_2\} = \mathcal{F}$ ) $$ \nabla_{e_1}e_1\equiv0\ ,\quad\nabla_{e_1}e_2\equiv c_4e_3, \quad \nabla_{e_2}e_1\equiv-c_4e_3\ ,\quad\nabla_{e_2}e_2\equiv 0 . \mod \mathcal{E} $$ 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.

It is not clear that there are any other examples in dimension $4$.
show/hide this revision's text 1

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 $4$.