The field strength is the force on a unit charge, so the field strength at the surface of sphere 1 is:
$$ F_1 = \frac{1}{4\pi\epsilon_0} \frac{Q_1 . 1}{r_1^2} $$
and the field strength at the surface of the second sphere is:
$$ F_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_2 . 1}{r_2^2} $$
Lets take the ratio $F_1/F_2$ to see which is greater. The constants cancel to give us:
$$ \frac{F_1}{F_2} = \frac{\frac{Q_1}{r_1^2}}{\frac{Q_2}{r_2^2}} $$
and I'm going to rewrite this slightly to make it obvious how you use your equality $Q_1/r_1 = Q_2/r_2$:
$$ \frac{F_1}{F_2} = \frac{\frac{1}{r_1}\frac{Q_1}{r_1}}{\frac{1}{r_2}\frac{Q_2}{r_2}} $$
Because $Q_1/r_1 = Q_2/r_2$ we can cancel them on the top and bottom of the fraction and we're left with:
$$ \frac{F_1}{F_2} = \frac{r_2}{r_1} $$
and because $r_2 < r_1$ this means the field strength at the surface of sphere 2 is greater than at the surface of sphere 1.
Trivially, you could just have the field be uniform in a finite region around the dipole and not uniform elsewhere, so that the electric field as a whole technically isn't uniform, but this might not be the spirit of the question you're asking. Fortunately, you can just as easily construct situations in which:
- the electric field is non-uniform and smooth, and
- there is at least one point where an electric dipole will simultaneously experience no torque and no force.
The torque $\vec{\tau}$ on the dipole is given by:
$$\vec{\tau}=\vec{p}\times\vec{E}$$
where $\vec{p}$ is the electric dipole moment vector. Likewise, the force $\vec{F}$ on the dipole is given by:
$$\vec{F}=\vec{p}\cdot\nabla\vec{E}$$
To enforce zero torque, we need only require that $\vec{p}$ and $\vec{E}$ are parallel at the position of the dipole.
For simplicity's sake, let's say that $\vec{E}$ points in the same direction everywhere, and that $\vec{p}$ is parallel to it. Let's call that direction the $\hat{x}$ direction. In other words, let's say that $\vec{E}=E(\vec{r})\hat{x}$ and $\vec{p}=p\hat{x}$. Then we have that
$$\vec{\tau}=0$$
by construction, and
$$\vec{F}=p\frac{\partial E(\vec{r})}{\partial x}$$
by virtue of everything pointing in the same direction. Wherever $\frac{\partial E(\vec{r})}{\partial x}=0$, a dipole will experience both zero force and zero torque.
For example, the field given by $\vec{E}(\vec{r})=kx^3\hat{x}$ has $\frac{\partial E}{\partial x}=3kx^2$, so a dipole will experience no force and no torque when $x=0$.
Best Answer
Start with the spheres separated by a displacement $d\hat{x}$ in an electric field $E\hat{x}$
If the spheres are very far apart, (compared to their radius), then before any charge moves, they will be separated by a voltage $V_0=Ed$. To compensate this, a charge $Q$ will flow from one sphere to the other, which changes their voltage by $Q/4\pi\epsilon_0R$ where $R$ is the sphere radius. So to make the voltages of both spheres equal, they have charges of: $$ Q=\pm4\pi\epsilon_0EdR/2 $$ Thus each sphere experiences a force from the electric field of $$\pm4\pi\epsilon_0E^2dR/2,$$ but they experience a force from one another of $$ \frac{1}{4\pi\epsilon_0}\frac{(4\pi\epsilon_0EdR/2)^2}{d^2} $$ Thus the force between the two spheres is smaller by a factor of $d/R$, which was defined to be big, and the spheres separate. They experience a much bigger force from the electric field separating them than from eachother.
It's certain that if they are separated initially by some distance greater than their radius, they are pulled apart. But obviously I have neglected to higher orders in approximation; one would need to do the following (in no particular order because some of these change order as you change parameters in the system):
I suspect that none of these considerations, no matter what distance scale, will change the underlying conclusion that the spheres separate. This is because having the spheres as separate as possible minimizes potential energy in the form of $\int E^2dV$. These conductors are eliminating the presence of electric fields near themselves, and by spreading out as much as possible, they maximize the volume over which they are counteracting electric fields.
Kind of interesting then that in the presence of an external electric field, a conductor with some flexibility will try to expand as much as possible. I didn't know that before. Recall also my comment to the question - even if the spheres are perpendicular to the electric field, they both get dipole moments in the same direction, making them push eachother apart (and the $\int E^2dV$ argument also still applies).