Dipole moment of conducting sphere in uniform electric field As stated in the title, uniform $E$-field generates dipole moment.
$$\vec p=4πε_0R^3\vec E.$$ 
One way to show this is considering the following configuration:  two spheres with radius R with uniform volume
charge density. One sphere has volume charge density $+ρ$, one sphere has volume charge density
$−ρ$, the center of the two spheres are at the distance $a$ ($a < R$) from each other.
The electric field caused by the two sphere at a point inside the overlapping region of the two
spheres is: $\vec E=\frac{ρ}{3ε_0}\vec a$  i.e. uniform.
So, uniform external electric field perfectly cancel out satisfying the condition that electric field inside the conductor is $0$.
What I wonder here is whether it is okay to fix the charge configuration only because of the above discussion. Of course, I know the result is true since mathematical approach using laplace equations and legendre polynomials gives same result. But I prefer creative approach like the above and want to understand it perfectly. 
I thought of uniqueness theorem but wikipedia says that 

The electromagnetism uniqueness theorem states that providing boundary conditions for Maxwell's equations uniquely fixes a solution for those equations. However, this theorem must not be misunderstood as that providing boundary conditions (or the field solution itself) uniquely fixes a source distribution. One counterexample is that the field outside a uniformly charged sphere may also be produced by a point charge placed at the center of the sphere instead, i.e. the source needed to produce such field is not unique.

Is there some type of uniqueness theorem to determine this situation? or other method?
 A: This is a bit of an old question but I stumbled upon it and I believe it a great learning experience and quite a beautiful question.
What you got right (most of it)
I'll begin by pointing out that you are entirely right! your configuration creates a perfectly uniform field in the ball!
(I'll use ball to refer to the volume, while sphere would be used for the thin shell. I highly recommend using those definitions to avoid confusion)
As you pointed out the uniqueness theorem doesn't mean that your solution is unique. In fact there are infinite configurations of charges that would create uniform field. Unfortunately your solution is slightly problematic, because you described your charges as an nonuniform distribution of charge inside the volume of a perfect conductor.
Why is that a problem? Because inside of a perfectly conducting body there are no electric fields! (Any field the body could be subjected to would create surface currents that would immediately cancel the fields for the volume of the body.) Without electric fields, why would the charge ever arrange itself so nicely?
As there are no way to create charge distribution inside perfect conductors we generally treat those solutions as "Non-physical", meaning mathematically correct but not real. Surface charges are on the other hand completely possible!
I will note, that your solution is on the right track. By taking the right limit and changing your notation a bit you can get the charge distribution as a surface charge.
So, now that we understand the problem, let's answer each of your questions
Is there a unique solution?
Or as you placed it "Is there some type of uniqueness theorem to determine this situation?"
I'm going to take some liberties and assume the conductive sphere is neutrally charged, otherwise there are infinite solutions. (differing only by the charge of the sphere) I'll also assume all the charge is on the sphere and not on the ball.
With those assumptions the solution is unique! Although I'm not aware of any named theorem to the problem, there is an easy way to see that there is only a single solution that has only surface charges.
We'll prove it thusly:

lets assume we have two distributions that create the uniform field $\vec E=A\cdot \hat z $ the solutions: $\sigma_1 (u,v)$ and $\sigma_2 (u,v)$. Both of them defined over the same closed, orientable surface.

($u$ and $v$ are parameters that describe our surface. With a sphere we could choose for example $\theta$ and $\phi$. $A$ is an arbitrary constant and $\hat z$ is an arbitrary direction)

Lets examine $\sigma_1 - \sigma_2 (u,v)$. From superposition we can immediately tell that the field inside the volume is zero. What is the field outside the body? Zero. Let's prove it.

You can get the solution immediately here by making an argument about the energy of the new electric configuration, which must be zero. We'll make the same argument in the proof, albeit a bit more formally.

Let's take a small ball around each part of surface, so it is small enough to not touch another part of the surface (there is only one surface connectivity group inside the ball). We'll use Gauss's law to find the charge inside the sphere. If it is zero everywhere then we have no charge and therefore no field and we are done. If there is a some area with positive charge then there must at least a single direction in which the electric field is negative. Let's imagine a "test charge" and move it with the field direction. There are three options:

*

*We will continue infinitely away from the body. But then it means the body has positive potential. In contradiction to the assumption that the body is neutral.

*We will reach a minimum of electric field somewhere in space. But a minimum of electric field is simply a point of negative charge! In our scenario all charges are on the surface. so this is also a contradiction.

*We will reach a minimum of electric field on the surface. But that means that we increased in potential! since $-\int \vec E \cdot \delta L=\Delta V$. But all points on the conductor have the same potential! Contradiction.

If the field is zero everywhere then there is no electric field anywhere and so no electric charge anywhere. Formally, due to Gauss's Law every possible volume would have zero divergence of electric field and thus zero average charge.
$\sigma_1 - \sigma_2 (u,v)=0 \Rightarrow \sigma_1 = \sigma_2$

and so we proved that $\sigma$ is unique.
What is the right solution?
You are already on the right track! If we take the limit of $lim{a\rightarrow 0^+}$ and increase $rho$ accordingly so to keep $\rho \cdot a$ constant (meaning our polarity is the same as before) we can immediately get the surface charges.
Let's look at an arbitrary slice of the ball, at height H from the top, we'll consider the top half for now. (Up is $\hat a$ for simplicity) The radius of this slice is $\sqrt{R^2 - H^2}$ the area of this slice is therefore the square of that times $\pi$ meaning $A(H)= \pi \cdot (R^2 - H^2) $.
What is the total charge at a single slice? $Q (H) = \rho (A(H)-A(H+a))$. All the charge is distributed over the thin shell, since it's simple to prove that the charge is zero anywhere else. We immediately get that $\sigma (H) = \rho (A(H)-A(H+a))/{2 \cdot \pi  \cdot \sqrt{R^2 - H^2}}$
$\sigma (H) = \rho (2H \cdot a)/{2 \cdot \pi  \cdot \sqrt{R^2 - H^2}}$
We can immediately find $(\rho \cdot a)$ since we know the polarity of the final charge distribution.
good luck
