Does a pair of disjoint homotopic cycles in the dual separate the graph?
Yes.  Let me write $\Sigma$ for the surface on which $G$ and $G^*$ are embedded.

Because the cycles $C_1$ and $C_2$ are homotopic, they are also in the same $\mathbb{Z}_2$-homology class.  So by definition, the symmetric difference $C_1\oplus C_2$ is the boundary of the union of some subset of faces of $G^*$; call this union of faces $U$.  (In fact, either $U$ or its complement $\Sigma\setminus U$ must be an annulus, but this isn't important.)

Because $C_1$ and $C_2$ are disjoint, the symmetric difference $C_1\oplus C_2$ is equal to the union $C_1\cup C_2$.  In particular, we have $C_1\oplus C_2\ne \varnothing$, which implies that both $U$ and its complement $\Sigma\setminus U$ are non-empty.  In other words, the subsurface $\Sigma \setminus (C_1\cup C_2)$ is disconnected.

Any path in $G$ can be viewed as a path in $\Sigma$ that avoids the vertices of $G^*$, and vice versa (up to homotopy).  Thus, the (graph) components of $G\setminus (E_1\cup E_2)$ correspond bijectively to the (surface) components of $\Sigma \setminus (C_1\cup C_2)$.  We conclude that $G\setminus (E_1\cup E_2)$ is disconnected.

The assumption that $\Sigma$ is orientable is never used.