Actually, the astronaut would only float completely free in the middle of the space station. Elsewhere, he will stick slightly to whatever side is closer to the Earth than is the middle, or farther from the Earth than is the middle. The reason is the tidal force from the Earth, which will be very small but probably detectable. If the acceleration from gravity is $g = GM/r^2$, then the tidal acceleration will be the derivative of this, namely $dg/dr = -2GM/r^3$. so the maximum tidal acceleration that the astronaut will feel, at distance $D/2$ from the center of the space station, where $D$ is the linear size of the space station, will be $GMD/r^3$. This is basically equal to $gD/r$.
Putting in some typical numbers, a 100 kg by weight astronaut in a space station of size $D = 10 m$, and $r = 6500 km$, we get a maximum tidal force of about $150$ milligrams weight!
The situation would of course be completely different in orbit around a neutron star, where an astronaut would float free at the center of a (very strong) space station, but would get crushed beyond recognition were they to venture away from the middle, as has been pointed out in various science fiction stories. The reason is that $M/r^3$ would be about $10^{14}$ times greater than for an orbit around the Earth.