AC-coupling capacitor destroys output signal?
Without the output capacitor the top transistor supplies all the current to the 8 ohm load. It is an emitter follower and this configuration is ideal for this type of drive. The lower transistor hardly plays any role in this circuit and you could actually remove it and get virtually identical results.
When you use a capacitor in series with the 8 ohm load, things change considerably. Now the lower transistor is expected to remove charge from said capacitor when the input signal is at its trough or most negative peak. However, the current sinking capabilities of the lower transistor (configured as a weak current sink) means that it can't do this to any degree comparable with the top emitter follower transistor's ability to source current.