I believe it will draw pretty close to locked rotor current (same magnitude as normal start), but only for a longer time than a normal start.
There are two odd-ball factors that may alter that situation slightly.
1 - locked rotor current can be obtained from equivalent circuit using s=1. If you solve using s=1.1 corresponding to 10% backward speed I think you will get a number slightly higher. I believe this number is valid but the difference is small, especially when reverse speed is slow.
2 - residual magnetism in the rotor may create some stator voltage in the spinning motor. If that voltage opposes the applied voltage at the moment of energization it seems to me there may be slightly different transient of the exponentially-decaying dc offset component than normal (possibly higher dc component than normal).
I think both these effects if they exist at all are small.