This circuit does not intentionally saturate the output transistors, as the filtering of C2 prevents the output from following the PWM input waveform. This is probably appropriate, since most brushless DC motors get confused by interrupted (PWM) DC.
Your output stage is a power operational amplifier that attempts to match the voltage across the motor to the ((10k+33k)/(10k)) filtered PWM input, which is 0-3V. That means that the motor voltage in this circuit can try to be 0-(4.3*3)V, or 0-12.9V (bounded by the 12V supply). The remainder is dissipated in the two output transistors which are in the linear mode. If the motor is 6W at 12V, this is 500mA. This is greater than twice the collector rating of the 2N3904, so even if they were OK in parallel, two is not enough. And at half output, if the motor needs say 300mA (it's not likely to be linear) this is 1.8W in the transistors. This is much more than double the rating of the transistor. Again, two is not enough.
You need a larger transistor, preferably a single one so you don't have to equalize emitter currents. I suggest any NPN of the TIP series transistors which you can get easily such as TIP31, with a small clip-on heat sink. The TO-220 package can theoretically handle 2W without a heat sink, but it can burn you or your board.
You may need other components on the collector of Q3 to prevent oscillation in the final stage, and unless you trust your PWM to be a stable 0-3V you might want to amplify and square it.