Twin paradox - observers counter orbiting Earth

Interesting question... with the assumptions you've specified, obviously observers B and C will have the same time on their clocks when they meet again because their situations are identical. Observer A, on the other hand, has to hold itself in place with a rocket or something, so it's not in an inertial reference frame. Based on that, A will have a different time on its clock when the three meet up again.

Let's see what the math says. Given a spherical, nonrotating Earth, we can use the Schwarzschild metric to describe spacetime outside it.

$$c^2\mathrm{d}\tau^2 = \biggl(1 - \frac{r_s}{r}\biggr)c^2\mathrm{d}t^2 - \biggl(1 - \frac{r_s}{r}\biggr)^{-1}\mathrm{d}r^2 - r^2(\mathrm{d}\theta^2 + \sin^2\theta\,\mathrm{d}\phi^2)$$

For the spacetime trajectories of the three observers, $\mathrm{d}r = 0$ (because they stay at a constant radius) and $\mathrm{d}\phi = 0$ (because they orbit in a single plane). So a bit of algebra gets you to

$$\frac{\mathrm{d}\tau}{\mathrm{d}t} = \sqrt{\biggl(1 - \frac{r_s}{r}\biggr) - \frac{r^2}{c^2}\biggl(\frac{\mathrm{d}\theta}{\mathrm{d}t}\biggr)^2}$$

In this formula, $r$ is equal to the radial coordinate of the three observers, $c$ is the speed of light, and $r_s$ is the Schwarzschild radius of the Earth. All three of those are constants. The only thing that differs from one observer to another is the coordinate angular velocity $\frac{\mathrm{d}\theta}{\mathrm{d}t}$, which is something like the angular velocity as measured by a distant observer. For observer B, this will be equal to some constant $\omega$, for C it'll be equal to $-\omega$, and for A it will be zero. This means that $\frac{\mathrm{d}\tau}{\mathrm{d}t}$ is equal for B and C, and slightly greater for A.

Now, this quantity $\frac{\mathrm{d}\tau}{\mathrm{d}t}$ is the rate at which proper time ($\tau$) elapses relative to coordinate time ($t$). The coordinate time is, again, basically what would be measured by a distant observer. So each time the three observers A,B,C meet up, the meeting takes place at the same coordinate time for all three of them. However, the proper time $\tau$, which is the time each observer measures internally, is not the same for all three. The fact that $\frac{\mathrm{d}\tau}{\mathrm{d}t}$ is larger for observer A means that for a given amount of coordinate time (like, say, the interval between two successive meetings of the three observers), A will experience more time than B or C. So if the observers start with synchronized clocks, when they next meet up, A will find that its clock is a little bit ahead of B's and C's clocks.

If you're curious about the numbers: we may not have enough precision to get an accurate result, but I can do this just to show how the calculation would work. Let's plug in the Earth's Schwarzschild radius of $r_s = 8.9\text{ mm}$ and the orbital radius of, say, the International Space Station at $r = 6750\text{ km}$ (rough average). We can also use the ISS's orbital speed of $r\frac{\mathrm{d}\theta}{\mathrm{d}t} = 7.68\ \mathrm{\frac{km}{s}}$ for observers B and C. That gives the following rates:

$$\begin{align}\frac{\mathrm{d}\tau_A}{\mathrm{d}t} &= 1 - 1.2027\times 10^{-8} & \frac{\mathrm{d}\tau_{B,C}}{\mathrm{d}t} &= 1 - 1.2355\times 10^{-8}\end{align}$$

The difference works out to $3\times 10^{-10}$. So over a 90-minute orbital period, the clock on A would come out ahead of that on B or C by 1.7 microseconds. But again, I'm not sure this number is necessarily trustworthy because we are talking about very small numbers here, and some of the GR effects I've neglected may contribute.