According to the General Theory of Relativity, the coordinate time distance per spacetime distance traveled by a particle freely falling into a black hole gets closer and closer to $0$ as the particle approaches the event horizon
$$\mathrm{d}s^2 ~=~ -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \left( 1-\frac{2M}{r} \right)^{-1}\mathrm{d}r^2 + \dots.$$
Equivalently, an observer looking from the outside will never see the particle cross the event horizon, even if he/she looked for an arbitrarily long time. However, Hawking radiation implies that black holes don't last forever, but instead shrink and fade away? What happens to everything that is "currently" falling into the black hole during that time?
Edit: What this seems to imply is that as a particle gets close to the event horizon, the rate and intensity of Hawking radiation will increase and increase until most of the mass of the black hole is ejected, in the form of Hawking radiation, out towards the falling particle.
Best Answer
If, as is canonically accepted, the black hole evaporates within a finite time, then the freely falling particle will be released from its gravitational attraction. By the premise of the question, in the observer frame the particle can not have passed the event horizon during the intervening period. This would also solve the quantum information paradox, as the information represented by the particle would never have been lost.