Settle in and get a drink. This will take a while.
First, how does a boomerang work? It's all in the wrist. That is, rotation and torques.
Let's ignore the fact that, generally speaking, a boomerang does not follow a horizontal circle. That's for later. Furthermore, in order to make things clearer, let's pretend that a boomerang is a helicopter. In addition to the rotating rotor disk, it has a body with the intuitive orientations. And let's say that throwing a boomerang (right handed, flat side of the boomerang to the right) amounts to launching the helicopter nose first but rotated 90 degrees CCW so that the rotor disk is vertical.
Two odd things happen to the ship, and both of them are precessions. Calling the right side of the rotor disk the advancing side and the left the retreating side, it's clear that overall the advancing side has greater velocity wrt still air than the retreating side, due to the aircraft velocity. This means that the advancing side produces more lift than the retreating side, and the result is a torque that tries to rotate the ship CCW (right side up, left side down). But here's the thing: this is a gyroscope, and it won't move that way. Instead, the system will precess, and the result will be that the nose will try to pitch up (in the aircraft's body coordinates). This will cause the boomerang to circle to the left, and that's why it goes in a circle.
But another odd thing (precession) occurs, and the cause of that is not settled. The common explanation is the rotors in the front half of the disk (leading) encounter undisturbed air, while the rotors in the back half (trailing) encounter disturbed air in the wakes of the rotors in the leading half. This produces a net torque which tries to pitch the nose up. Another explanation is that the leading elements of the rotor disk interact differently with the translational air flow differently than the trailing edges, without taking wake disturbances into account, and the result once again is a differential lift which produces a torque which tries to pitch the nose up. Both arguments seem reasonable, so I suspect both are correct, but it's not at all clear which effect dominates. However, either way the torque is produced, the result is the same. Just as the boomerang cannot respond directly to the advancing/retreating torque, neither can it respond to this torque. The result is a second precession, this one rolling the system CW (again in body coordinates). In this case, this works out beautifully for real boomerangs: as drag slows the arms, the system rolls right, and the total lift becomes more vertical, so the boomerang does not crash into the ground. For a perfect throw, the boomerang comes around slowly rolling until it is horizontal just as it returns to the throwing point. Furthermore, the "nose" is pitched up slightly, braking it. As the forward speed drops, so do the induced torques and associated attitude angular rates. If the torques are proportional to velocity, the boomerang maintains a perfect circle. At some point the forward speed drops to zero, but the arms are rotating fast enough for the thing to hover, then gently start to fall as the arms slow. If you do it right, it's quite remarkable and easy to catch.
So, with that out of the way, what happens in microgravity. Well, that depends. First, let's ignore drag losses and assume both rotational rate and translational velocity remain the same. Then, depending on rate and velocity and a whole bunch of other stuff (air density, airfoil performance, mass, dimensions, etc) the 'rang will precess at two rates, P (pitch) and R (roll). For a perfect circle on the ground, the first precession rate P is exactly 4 times the second, R. In other words, when the boomerang has made a perfect 360 circle, it has also rotated 90 degrees to the horizontal, so P = 4R.
From this, it's clear that a circle will take T seconds, where $$T = \frac{2\pi}{\sqrt{{P^2}+{R^2}}}=\frac{2\pi}{\sqrt{5R}}seconds$$ and the "vertical" displacement D of the start and top points will be $$D =\frac{V\sqrt{5R}}{8\pi}$$ At the end of the first loop, the plane of the boomerang will be rotated 90 degrees, so the total path will form a closed loop with 4 segments of helix, with the start/stop points of each segment forming a square. In general for differing ratios of P and R I believe you'll get a series of helices whose central axis itself forms a circle, with the boomerang path lying within the surface of a toroid.
Ah, but that's in an ideal world. What about frictional losses? That gets tricky, and is beyond what I'm willing to spend the time on. Presumably P and R do not respond identically to changes in V, for instance. The important factor is the relationship between rotational drag (how fast the boomerang stops spinning) and translational drag (how fast the boomerang slows down). Because the boomerang is always "pitching up" the overall disk is constantly at a positive angle of attack and will suffer considerable losses. Obviously, the end point is a stationary boomerang, but the end game depends on which slows more quickly, translation or rotation. If translation dies first, as in the "perfect throw" the boomerang moves perpendicular to its plane of rotation due to lift from the arms. If the rotation dies first, the boomerang is no longer gyroscopically stabilized and will tumble in a chaotic fashion. In general, you'd expect the transition from helix to endgame to be very hard to manage, since it's very unlikely the various rates, velocities, torques and drags are simply and linearly related.
