(Ah, a very old question which is not yet answered... and I don't find a duplicate answer, either, though the problem is very nice ...
the comment is point you there, and that the user "centralcharge" has edited the question is amazingly fitting, too :) but it's not yet elaborated)
Yes, you can use the method of images. Normally you are right, it requires the object to be grounded, but for a sphere there is a nice "trick".
You have to first find an image, that would make the potential on the sphere zero. How to do this is apparently clear to you. This would solve the grounded case.
But there is no requirement for the potential to be zero. It has just to be equal on the sphere. And this requirement is not violated, if you put some charge in the center of the sphere. This central charge will just shift the potential on the sphere, but obviously not break spherical symmetry.
Take the extra charge just as big as you need to adjust the right net charge. Then you have solved the problem for non-grounded sphere.
So now to your full problem: In what distance would the sphere with $+Q$ net charge and the point charge of $+Q$ be at equillibrium?
You can solve this with an quite complicated equation or just guess the solution and check. Let's guess the golden ratio: $r = \frac{\sqrt 5 + 1}{2}R$, and briefly check this:
I'll set the radius of the sphere and the charges to $1$. Now assume the outside charge has distance $\phi = \frac{\sqrt 5 + 1}{2}$ from the center of the sphere. Then the induced charge is $-1/\phi$ and it's distance from the center is $1/\phi$ too. The charge in the middle has to be $1+1/\phi$ for the net charge to be $1$
Now it holds $1/\phi = \phi-1$, that's why $\phi$ is nice. It follows:
The force between the outside charge and the induced charge: the distance is $1$, the product of charges is $1/\phi$, so the force is an attraction of $\frac{1/\phi}{1^2} = 1/\phi$.
The force between the outside charge and the central charge: the distance is $\phi$, the product of charges is $\phi$, so the force is a repulsion of $\frac\phi{\phi^2} = 1/\phi$.
The forces are equal, qed.
(...if you solve the equations you see, that this is indeed the only real solution greater than 1)
If the sphere is initially uncharged, then the electric flux through its surface is zero, by Gauss' law. If we add just one point charge $q'$, then we will find a net flux through the surface, which is wrong. However, add a second charge $q^{''}=-q'$, then the net flux is given by the enclosed charge $q'+q^{''}=0$, and all is well.
Best Answer
Your answer would have been correct if, for example, the spheres were non-conducting and if the charges were distributed uniformly over their surfaces.
However, since the spheres are conducting, the surface charge distribution on each sphere will be altered because of the repulsion from the charges on the other sphere. In particular, the charges on each sphere will be pushed away by the charges on the other sphere. This will cause the charges on opposite spheres to be further away from each other, and the force of repulsion to be less than in the case of a uniform surface charge distribution.
Addendum 1. Why would the force be the same if the spheres were non-conducting with uniform surface charge densities? Well, without going into mathematical detail, note that using Gauss's law on each sphere would show that the electric field outside of that sphere would be the field of a point charge with the same total charge. Therefore, each sphere would simply "see" the other sphere as a point charge, and the result follows.
Addendum 2. Let's be a bit more precise in showing that when the spheres are conducting, the force will be lower than in the point charge case. By addendum 1 above, we know that the force would be the same of the spheres had uniform surface charge distribution with total charge equal to that of each point charge. Therefore, it suffices to show that the the surface charge distributions in the conducting case would lead to a smaller force.
To show this, divide each sphere into a large number of charge elements, and suppose that the charge distributions on the surfaces of the conducting spheres begin uniform. How exactly do the charge elements rearrange themselves on each sphere because of the influence of the other sphere? Well, note that if the other sphere were not there, then the charges would stay put and the distributions would remain uniform. Since the other sphere is there, the charges feel a force that pushes them away from the other sphere. In particular, picking any pair of charge elements from opposite spheres, we see that the distance between such a pair will be larger in the new surface charge distribution than in the uniform case. In particular, if we perform a sum over all pairs of surface charge elements on opposite spheres of the coulomb force between them, then every term in the sum will be smaller than in the uniform case, so the total force will also be smaller.