Yes, the equation is indeed "something like the integral of E2 "
To be precise, it is
W=(ε0/2) ∫all space E2 dV
The electric field described by the images only holds wherever the boundary conditions are satisfied by both the original configuration AND your simpler image charges scenario. In the case of a grounded conductor that region is above the plane, on the side of the point charge q. Assuming the the plane is at z=0, the field at z<0 is:
- Zero in the case of the grounded plane and charge
- Perfectly symmetrical (dipole field) in the case of the image charges
The z<0 region falls outside the boundary condition-satisfying region. The field E is not the same in both cases here
So, just intuitively, without solving the integral, since the E-field exists only in half of all space with the conducting plane, and in all space with the image charge, it should be fairly obvious to you why the energy "in the metal plate case is half that of the energy in the 2 point charge case".
In answer to your secondary question, this kind of problem usually involves an infinite plate. If it isn't infinite, then you have edge conditions and all the standard problems that realistic geometries bring with them to deal with. So we assume it is infinite. That implies that it is grounded, even if not specifically mentioned. The plate goes to infinity=> potential=0 at infinity=> plate is an equipotential surface=> plate is grounded
Hope that helped!