Driving MOSFET from MCU with BJT
D1 isolates the base drive of Q2 from the emitter output, so when Q1 turns on it pulls the base of Q2 to -0.7V relative to the emitter and turns it off before pulling the output down. 
Yes, you could add a diode, however the drive will be reduced a bit by the lowered voltage. You can reduce the resistor value to compensate. 
This totally depends on the drive voltage. Right now the base current is (Vi-0.7V)/1K mA. With a diode it will be more like (Vi-1.4V)/R2. So if Vi is 3.1V (say output voltage of an MCU with 3.3V supply) you have 2.4mA drive, and R2 should be more like 700 ohms (you could use 680, for example). 


That's assuming you're using transistors, however you drop a bit of a bombshell at the end- TIP122 not a transistor- it is an NPN darlington with resistors, a really slow one- switching times in microseconds. It does not fit your schematic in the question nor is it really appropriate for this application.