As I said, it's valid for any shape of the conductor and any charge distribution: As long as there's no charge inside and the situation is static, there's no electric field inside the conductor ("Faraday Cage").
For the sphere you can even evaluate the Green's function explicitly with help of the image-charge method. The Green's function G(\vec{x},\vec{y}) is given by the solution for the situation, where a unit charge sit's at \vec{y}. It's clear that the spherical shell must be an equipotential surface. The boundary conditions tell us that the tangential component of \vec{E} must be continuous, and thus the potential inside the whole sphere is constant (i.e., \vec{E}=0 inside the sphere). The field outside is given by the original unit charge at \vec{y} plus the field of an image charge inside, located at along the direction \hat{y} (the center of the sphere is in the origin of the coordinate frame) the sphere such that the sphere becomes an equipotential surface.
Then one has to think about the total charge on the surface of the sphere. If the sphere is grounded, this influenced charge is given by the mirror charge. If the sphere is neutral and insulated, one has to add additionally the Coulomb field of the oposite image charge in the center.
Thus, outside the sphere one has a relatively simple superposition of the field of the original unit charge and one or two image charges inside the sphere. In reality, as argued above, there are no charges inside the sphere, and thus the interior is field free.
You find the details of this calculation in any standard textbook on electrodynamics, e.g., Jackson, Classical Electrodynamics.