Imagine a graph of a sine wave, centered around 0 on the y axis. If the signal goes up to 1 on the y axis and down to -1, you have a 2V peak-to-peak signal. (Which is not the same thing as 2V RMS, but let's ignore that for simplicity's sake.)
Now imagine that the signal is shifted up so the sine wave is centered around +.1 -- its upper peak is 1.1V, and the lower peak is -0.9V. If this were a voltage, it would still be 2VAC p-p, but with a 0.1V DC offset.
In the context of a split power supply like in a CMoy or a META42 amp, shifting the voltage means that the voltage potential from virtual ground to each rail is different. You might get +4V and -5V from a 9V battery, which would mean virtual ground has shifted by 0.5VDC.
The DC offset I think you're talking about, finleyville, is on the op-amp's output. If your virtual ground shifts, then you don't really get a of DC offset on the op-amp's output, because the headphones use the virtual ground, too -- as far as they are concerned, the signal is still centered around virtual ground. When we say there's a DC offset on the op-amp's output, it's relative to the virtual ground, whatever voltage potential it may have.
The problem with DC offsets on the op-amp's output is that your headphones work best when the output signal is precisely centered around the ground potential. The more DC offset there is, the poorer they perform, until you get the extreme case where the signal is entirely positive (or negative) relative to ground, in which case the driver is pushed in one direction constantly -- it doesn't move back and forth, and thus doesn't produce sound. It may even be damaged.
With FET-input op-amps, you have to work pretty hard to get more than about 20mV of DC offset, which is completely ignorable in most circumstances, given that the signal level is hundreds of millivolts at minimum. It's with bipolars, where offsets can be multiple volts, that you have to worry about DC offset.