Curved space 0 - this is easy

In the first approximation the balls will have the same velocity and orbital radius as the spacecraft. They will move with the same velocity on the same circular orbit. As they move together with the space station, their $x,y,z$ position does not change. Their distance does not change. The answer is 2 m. Anyone getting that is essentially correct.

The story is a bit more complicated.

First, to be exact, the distance from the center of the Earth will be a bit larger for the two balls than for the space craft $R'=\sqrt{R^2+\Delta x^2}$ , where $\Delta x = \pm 1.0$ m. That will cause the balls to move (slowly) relative to the spacecraft. Also, if the distance from the center of the orbit is changed, there is no guarantee that the orbit remains circular. The two balls may actually move a little bit relative to each other. But we can be sure that in one full period they will be back at the same distance.

This is not all: The balls have an initial velocity that is slightly different. For the main window of the space station to point in the direction of Earth all the time, the space station must be rotating with an angular velocity of $\Omega= 2 \pi/T$ . If the ball in the forward position is at rest relative to the spacecraft, it will have a radial velocity of $\Omega \Delta x$ = 0.87mm/sec in the direction toward the Earth; the other ball will have the same radial velocity in the direction pointing away from the Earth. While this velocity is really small, in free space (with no gravity) each ball would move about 6.28 meters in opposite directions in 120 minutes.

This is not happening in the presence of gravity. The extra velocity will cause the balls to take up an elliptical orbit, but the semi-major axis $a$ of the two balls' orbit will be the same. To see this, let us express the semi-major axis with the energy: $a=-GMm/2E$ and, in first order in the small parameter $\Delta E$ , $\Delta a = GMm/2E^2 \Delta E = R \frac{\Delta E}{(-E)}$ . Here the small change in the energy is $\Delta E = \frac{1}{2} m\Omega ^2 \Delta x^2$ , independent of the sign of $\Delta x$ . According to Kepler's law, if the semi-major axis is the same, the period will be the same, too. Therefore, instead of moving away from each other, the two balls will move on orbits of the same period.

In summary, the two balls are not moving on exactly the same circular orbital and the astronaut will see them moving relative to each other. However, at the time of a full orbit is completed, they will come back to the same distance relative to each other, 2m. (The only possible problem is that the period of the balls is slightly different from 120 min, but the correction is second order in $\Delta x$ and the effect is very small.)

On the other hand the second order correction in the period will probably lead to a measurable shift of the balls relative to the space craft, similar to the effect caused by the change of the radial position mentioned at the beginning.

See my other problems, "Curved space 1", "Curved space 2", "Curved space 3" for more.