Can an accelerated spring in static equilibrium in its rest frame have different forces at each end? Let's say I have two  identical masses connected by a spring:

If F1 = F2 = F then the forces on either end of the spring are also F when the whole arrangement reaches static equilibrium. 
Now if F2 > F1, the whole arrangement accelerates eventually reaching static equilibrium in the instantaneous rest frame. But this implies to me that the force on the left end of the spring is F2, and that on the right is -F1 so that the two masses accelerate identically.
Is this correct?
 A: Let's say that the masses of the first and second objects are $m_1$ and $m_2$, respectively. The acceleration of the two masses will be $a=(F_2-F_1)/(m_1+m_2)$. So, first, the force on the first object due to the spring cannot be $F_2$. If it were equal to $F_2$ then the acceleration of the first object would be $(F_2-F_1)/m_1$ which is different from the acceleration calculated above for the coupled masses. When the masses reach a steady-state situation (or what you call "static equilibrium" then $m1$, $m2$, and the spring all have to be moving together with the same acceleration or else there will be net movement between them.
So what is the force acting on $m1$ due to the spring? Well, we know that the acceleration of $m1$ has to be equal to the acceleration calculated above for both masses, so we have the equation: $(F_x-F_1)/m_1 = (F_2-F_1)/(m_1+m_2)$ where $F_x$ is the force on $m1$ due to the spring. So $F_x = F_1+(m_1/(m_1+m_2))(F_2-F_1)$. Note that this is not equal to $F_2$. Similarly, if $F_y$ is the force on $m2$ due to the spring, then $F_y = F_2-(m_2/(m_1+m_2))(F_2-F_1)$. A little algebra can show that actually $F_x=F_y$. So the answer is that the (massless) spring exerts forces of the same magnitude (but opposite directions) on both $m_1$ and $m_2$. For the special case that $m_1=m_2$, you can see from the above equations that $F_x$ (or, equivalently, $F_y$) is simply the average of the magnitudes of the two forces $F_1$ and $F_2$.

A: 
Your assumption that the tension forces in the spring are equal to the applied forces ($T_1=F_2$ and $T_2=F_1$ where $F_1 \ne F_2$) leads to contradictions.
The resultant force on each of the 2 masses would then be the same ($F_2-F_1$ to the right, because $F_2 \gt F_1$), and both would accelerate to the right. If $m_1=m_2$ then they would have the same acceleration, as required for them to be at relative rest. However, if $m_1 \ne m_2$ then they would have different accelerations and could not be at relative rest.
The forces $T_1, T_2$ acting on the spring are equal and opposite to those acting on the masses. The resultant force on the spring would then be $T_1-T_2=F_2-F_1$ to the left. This means that the spring will accelerate to the left while the masses accelerate to the right. This is not possible. 
Furthermore, if the spring is massless ($m_3=0$) then its acceleration would be infinite for all non-zero values of $F_2-F_1$. Again, this is not possible. (If $F_2=F_1$ then there would be no acceleration for either of the masses or the spring.)
Conclusions : (1) If $m_1 \ne m_2$ then we cannot have $T_1=F_2$ and $T_2=F_1$ even if $m_3 \ne 0$. (2) If the spring is ideal ($m_3=0$) then we must have $T_1=T_2$. 

The General Case
It is required that the spring and 2 masses have the same acceleration $$a=\frac{T_1-F_1}{m_1}=\frac{T_2-T_1}{m_3}=\frac{F_2-T_2}{m_2}=\frac{F_2-F_1}{m_1+m_2+m_3}$$ From these equations we deduce that $$T_1=F_1+m_1a=\frac{m_1F_2+(m_2+m_3)F_1}{m_1+m_2+m_3}$$ $$T_2=F_2-m_2a=\frac{(m_1+m_3)F_2+m_2F_1}{m_1+m_2+m_3}$$ Since $F_2 \ge F_1$ it can be seen (as expected) that $$F_1 \le T_1 \le T_2 \le F_2$$ Equality applies when $m_1, m_3, m_2=0$ respectively. In particular, if $m_3 \ne 0$ then $T_1 \lt T_2$. If the spring is massless ($m_3=0$) then as expected the tensions are the same at both ends : $$T_1=T_2=\frac{m_1F_2+m_2F_1}{m_1+m_2}$$
