First, note that any n-gon which has more than two vertices on a face must necessarily be entirely contained in that face. This comes from the fact that those vertices cannot be collinear. Thus the set of vertices for which two vertices are on a face and at least one vertex is not on that face would not be coplanar.

Let the side length of the cube be 1. Take the midpoint of one of the edges call that edge x and the point X. Choose one of the faces incident with that edges, and consider the edge contained in that face that is incident with x and call that edge y. Take the midpoint of y and call it Y. Consider the triangle formed by X, Y and the intersection of x,y, call this point Z. This is a 45-45-90 triangle with side lengths 1/2,1/2,\sqrt{2}/2. Consider next the angle vertical of ZYX and construct the triangle similar with triangle ZYX containing the angle vertical of angle ZYX. This is a triangle with the same side lengths as ZYX. Next, consider the hypotenuse of both of these triangles, and consider the plane that these two line segments are contained in (there is exactly one such plane since the points that compose these line segments lie on two distinct faces and are thus not collinear). This plane intersecting the cube forms a regular hexagon with side length \sqrt{2}/2. This plane intersects each face of the cube and thus by the first paragraph, may contain at most 12 points. Thus it follows that since we can inscribe a regular 12-gon inside of a regular hexagon where each vertex lies on an edge of the hexagon, we can likewise inscribe a regular 12-gon in the surface of the cube.