One specific case of your problem is to partition $G$ into $A$ and $B$ such that the induced subgraphs on those two vertex sets are trees. We will refer to this task as partitioning a graph into two trees. This is a specific case of your problem because maximizing the number of edges that are cut corresponds to minimizing the number of edges that are not cut, and given that $A$ and $B$ are connected, the minimum possible number of edges that are not cut is $(|A| - 1) + (|B| - 1) = |G| - 2$ in the case that both $A$ and $B$ are trees.
For a planar cubic graph $G'$, the dual $G$ of $G'$ can be partitioned into two trees if and only if $G'$ has a Hamiltonian cycle. In particular, if $H$ is a Hamiltonian cycle in $G'$, then the corresponding partition of the vertices of $G$, aka faces of $G'$, into the ones interior to $H$ and the ones exterior to $H$ is a partition into two trees.
Thus, if you further restrict your problem to the case of partitioning the dual of a planar cubic graph into two trees, the specific case you are left with is equivalent to finding a Hamiltonian Cycle in a planar cubic graph. Since that task is NP-hard, so is this special case of your problem. And if a special case of your problem is NP-hard, then your problem as a whole is also NP-hard.