When the point charge is not at the center of the sphere, the electric field lines will not intersect the sphere at right angles. Consequently, there is an initial component of electric field along the surface of a conductor. We know this results in a force on the charge carriers inside the conductor, and these charge carriers will re-arrange until the electric field is once again perpendicular to the conductor.
At that point you have a charge concentration (positive or negative, depending on the polarity of $q$ and whether you are looking at the point closest to $q$, or furthest from it). The polarization would result in a force on $q$ - attracted towards the point of the sphere that it's closest to, since the charge concentration there is greatest.
There is no conflict with the uniqueness theorem. The UT only states that "if the solution meets the boundary conditions, it is the solution". And the above describes such a solution - the boundary conditions (E perpendicular to conductor surface) are met because there is a redistribution of charge.
The following diagram illustrates this:
I drew the field lines inside the sphere as straight lines initially, then attempted to show how they have to bend in order to meet the conductor surface at right angles. The only way I can think of bending the field lines at B is to have a charge of the same polarity as q on the surface. There can be no net field on the dotted surface (inside the conducting shell).
So getting down to brass tacks: "can there be a region where the surface charge is positive" (assuming that $q$ is positive)?
This is a question being asked on this site - and I reproduce the diagram given (for the correct answer):
The reasoning follows from the two hints supplied:
The field inside the conductor must be zero. The charge inside the shell is off-center, and hence the charge on the inner surface of the shell will arrange itself asymmetrically to cancel the field of the large positive charge.
The field inside the conductor must be zero. To the charges on the outer surface, it is as if the inside of the conductor were completely neutral. Thus, the charges on the outer surface will feel no force other than their own mutual repulsion, and will therefore have no preferred direction.
Now if you think about this, you can see that at every point where the electric field lines from $q$ hit the surface, you need an opposite charge to cancel the electric field - otherwise you end up with an electric field inside the conductor. And we know that can't happen. In the diagram below, $E_1$ and $E_2$ must point in opposite directions to cancel inside the conductor.
Best Answer
First and foremost, I don't think you could ever prove Gauss's Law wrong. If it says the field (and hence force) is zero, it is indeed zero, and accordingly its consequences. Now if you want the physical situation that causes the force experienced inside to be zero, it is crudely represented in the diagram. Consider the major bowl and the minor bowl (I like calling them that!). While the major bowl has more charge (due to larger surface area), it is further away. In the case of minor bowl, the surface area (and hence charge) is lesser, but so is its distance. For both the bowls, the increase/decrease in charge cancels out the decrease/increase in the distance from the point charge, such that the resultant fields due to each bowl are equal, and opposite, which then cancel each other, and hence field(force) is zero. Similar argument for center or any other point inside the sphere may be given.
P.S. the lines represent the direction of force due to each elemental charge (repulsive, as both are negatively charged). I forgot to draw similar lines for the minor bowl (in the opposite direction).