Without the IR filter, the residual 808 and 1064nm light emitted from the laser would be negligible... maybe on the order of 10-20mW
The crystal set sits basically on top of the diode and blocks most, if not all of the light. So any residual IR you get out will 1) not be much power and 2) not be in a beam shape of any sort, because by the time it makes it out the end of the laser the light has bounced off so many things inside that it just kinda dumps out the end
Any IR leaking through the crystals will not be collimated like the green beam, it will be like a flashlight of IR.
And no, the amount of IR coming out of the crystals will not equal the difference between the pump diode and the green light coming out. The process is inefficient, so a lot of light is absorbed into the crystals. You may ahve noticed or read the the crystals can get very hot during operation; high-end lab units will even have a dedicated TEC for the crystals themselves to keep them cool. This heat also comes from the output of the pump diode. So in theory:
Pump diode output power = green light output power + IR leakage output power + power absorbed into the crystals that turns into heat. That power absorbed into the crystals, and lenses as well for that matter, is a very large chunk of the pump diode's total power.
For a 50mW green, I would guess maybe somewhere between 200 and 500mW of power from the pump diode. Someone could probably give a better number, especially if you could give the model or manufacturer of the laser.
So if I take out the crystals, the IR filter, other junk and maybe cut it open to make open can I end up with a powerful IR laser burner, but how much divergence should I get now with all the extra crap taken out?
I think Sam quoted it at 70-80% loss. Not very efficient, to say the least...
Regarding the OP's question, I would have thought that the diode, along with the OC mirror are the key components for coherent light and that the Nd:YVO4 & KTP are only used to alter the wavelength to 532nm... . I realize that this diagram from Sam's FAQ is simplistic, but it would suggest that removing the crystal and KTP layer will still leave you with a viable laser (yes, ok, maybe the lens needs to be repositioned)...