Hubble flow and inertial reference frames Looks like different questions, but they are basically the same:
1) Let us imagine a mostly empty expanding universe with constant $H$. Two observers (Ginger and Fred) that are separated by a cosmological distance but follow the Hubble flow will see each other accelerating (if measuring proper distances). Will each consider the other as non-inertial? What if Ginger has an electric charge: Will Fred see that this electric charge is accelerated but not radiating? 
2) And what about the reverse? Put two opposite charges at the distance at which they are at rest with respect to each other (I assume this distance must exist because if the charges are too close, they will strongly attract and move towards each other; but If they are far enough from each other, their attraction force will be weak and the charges will see each other moving away due to the Hubble law). Do this mean that the static force no longer follows the inverse square law if described in terms of the proper distance? What if we describe it in terms of the comoving distance? 
3) If the charges are "bounded" that way, it looks like neither particle will be in an inertial frame: Presumably the center of mass will move with the local hubble flow, and from its point of view  the particles are at a fixed distance, at rest relative to the CM, but accelerated relative to nearby particles that follow the hubble flow. Will the nearby particles see that the charges radiate? Will an observer at the CM see these charges at rest but radiating?
 A: The answer to (1) is fairly simple. GR allows us to consider any motion along a geodesic as inertial[citation needed]. Since Ginger and Fred are moving with the Hubble flow, they can recognize that they are both moving along geodesics and, thus are inertial, which also means that Fred would not see Ginger radiate.
The answer to number (2) is that the inverse square law still applies and that, in fact, there is no distance at which these charges will remain at rest w.r.t. each other. Keep in mind that the static force means acceleration is dependent on the inverse square of the separation. On the other hand, the recessional velocity from expansion means only that the velocity is proportional to the separation. The Hubble flow does not impart an acceleration based on distance to anything but a comoving observer. If you manage to find some distance where the particles are not moving farther away from each other, then the recessional velocity of the surrounding regions of space is not increasing. However, since they still feel a static force, they are experiencing an acceleration towards each other. This means that in the very next moment of time, the recessional velocity will be less than the velocity imparted rom the static force and they will begin moving together (at which point, the recessional velocity begins decreasing and they move together faster). The best you might be able to do in terms of finding an equilibrium (I say "might" because I don't feel like doing the math to see if it's possible) is to find some situation where the increase in recessional velocity and peculiar velocity keeps pace while the separation of the particles grows. That net effect would be to maintain the apparent increases in proper distance per unit time between the charges.
Let me rephrase that answer, just in case it wasn't completely clear. You're thinking of this as "either it is too close and the distance decreases or is too far and the distance increases". This is not the way to think about it. What you actually have is "either it is too close and the velocity increases inwards or it is too far and the velocity increases outwards". The difference? The equilibrium point is not where velocity is zero; it's where the velocity doesn't increase in either direction, but the nature of recessional velocity guarantees that this point will have a positive outwards velocity. That's the only way to counter an inwards acceleration.
So the inverse square law still applies. The force exerted on the particles is still proportional to $\frac{1}{r^2}$, you simply have a velocity offset that is proportional to separation that also needs to be taken into consideration.
Say you have a time step $\delta t$. The total velocity, $v$, is a combination of the recessional velocity for the separation, $R$, and peculiar velocity, $v_p$:
$$v(t)\propto HR(t)-(v_p(t-\delta t)+\frac{\delta t}{R(t-\delta t)^2})$$
Let's remember this relation just shows proportionality; I didn't include all the constants in the inverse square term. But basically, if you find a place where $v=0$, then the first term won't change in the next moment of time, but the second term will, guaranteed. This means the particles can never remain at rest w.r.t. each other for more than a moment, but you can clearly see the inverse square law affecting the only acceleration term in the expression.
I suppose this answers the question of whether or not an observer at the CoM will see the charges radiate. They aren't going to be at rest, so in all probability, yes they will radiate. Comoving observers near each particle will also see the particles radiate because they are accelerating. 
