Poor sound, large supply current and reduced reliability due to the high junction temperatures if the op-amps have good built-in current limiters, as they usually have. Smoke when they haven't.

I'll give you an example:

Imagine you make an amplifier with a gain of ten using two paralleled op-amps of the same type, but due to the inevitable exemplaric spread, they behave slightly differently. The load resistance of the amplifier is 100 ohm including its feedback network.

Op-amp 1: open-loop gain 100 000, offset 0 V, current limit at +/- 21 mA. To keep things simple, I will assume an open-loop output impedance of zero.

Op-amp 2: open-loop gain 100 000, offset 100 uV, current limit at +/- 20 mA.To keep things simple, I will assume an open-loop output impedance of zero.

When the amplifier has zero input voltage, op-amp 1 will try to make the output voltage of the whole amplifier 0, while op-amp 2 would like it to be +1 mV (ten times its 100 uV offset; I've neglected the effect of the finite open-loop gain, otherwise it would be +0.9999000099990000... mV). Op-amp 1 will win because it has the highest current limit. Op-amp 2 will source as much as it can source, the full 20 mA, but op-amp 1 will sink that current and keep the output at 0 V. The current through the load will be 0, so that has no impact.

When the amplifier has an input voltage of +20 mV, op-amp 1 will try to make the output voltage of the whole amplifier 200 mV, while op-amp 2 would like it to be 201 mV. It will end up in a state where the output voltage is 200 mV, the current through the load will be 2 mA, op-amp 2 sources its full 20 mA and op-amp 1 sinks the difference, that is, 18 mA.

So far everything behaves as a single-ended class A amplifier with op-amp 2 as the current source, but that's no longer the case with negative input signals.

When the amplifier has an input voltage of -20 mV, op-amp 1 will try to make the output voltage of the whole amplifier -200 mV, while op-amp 2 would like it to be -199 mV. It will end up in a state where the output voltage is -199 mV, the current through the load will be -1.99 mA, that is, 1.99 mA into the output of the amplifier, op-amp 1 sinks its full 21 mA and op-amp 2 sources the difference, that is, 19.01 mA.

So, besides the inefficiency and the reduced reliability due to the resulting high junction temperatures, you actually get a jump in output voltage at the point where the load current is such that the winning op-amp changes.