It is a standard variation of the Travelling Salesman Problem.
To force an edge to be a part of the answer, break it in half and insert a new vertex.
Update A reduction in the other direction consists of "breaking all vertices in two".
More precisely, given a directed graph $G$, form another graph $G'$. For each vertex $v$ in $G$, create a pair of vertices $v_i$, $v_o$ in $G'$. For each edge $(uv)$ in $G$, create an edge $(u_ov_i)$ in $G'$. Also, for each $v_i$, $v_o$ add an edge ($v_iv_o)$, and let it belong to $F$. Now every cycle in $G'$ that contains all edges in $F$ also contains all vertices, and corresponds to a Hamiltonian cycle in $G$.