This is inaccurate. The object can fall in. Where the argument fails is that it relies on the idea that an observer will see the object redshift forever. But that applies in the classical GR realm only. However, when combined with QFT (which is required for BH evaporation) it no longer holds.
In the classical approach, the light that the object emits a second before crossing the horizon will take years to reach the observer, at a millisecond it'll take decades, at a microsecond it'll take centuries, etc. But it's always still coming, never stops completely. So in the classical analysis, an observer will always see the object, unboundedly redshifted, infinitely near the horizon, but not quite there, forever.
However, in the presence of Hawking radiation, the black hole loses mass, and thus the horizon shrinks. When this happens, that last bit of light the object emitted before crossing the horizon will now reach the observer in finite time. In particular, the observer will see that long _before_ they see the black hole evaporate entirely; one photon's worth of radiation is enough to do the trick.
You're correct that the object will experience this time dilation by seeing more radiation as it gets closer to the horizon. It'll also start getting pelted from behind by infalling radiation. But nothing will blow up as it crosses the horizon, so it'll safely continue on to experience whatever quantum-graviational phenomena happens at/near the "singlularity".
In the classical approach, the light that the object emits a second before crossing the horizon will take years to reach the observer, at a millisecond it'll take decades, at a microsecond it'll take centuries, etc. But it's always still coming, never stops completely. So in the classical analysis, an observer will always see the object, unboundedly redshifted, infinitely near the horizon, but not quite there, forever.
However, in the presence of Hawking radiation, the black hole loses mass, and thus the horizon shrinks. When this happens, that last bit of light the object emitted before crossing the horizon will now reach the observer in finite time. In particular, the observer will see that long _before_ they see the black hole evaporate entirely; one photon's worth of radiation is enough to do the trick.
You're correct that the object will experience this time dilation by seeing more radiation as it gets closer to the horizon. It'll also start getting pelted from behind by infalling radiation. But nothing will blow up as it crosses the horizon, so it'll safely continue on to experience whatever quantum-graviational phenomena happens at/near the "singlularity".