Does resistor power rating mean 'real' power?
I have a 36 ohm resistor through which a current of 9.22mA @ 86º passes.


If the frequencies involved are not too high, the voltage associated with this current will also have a phase angle of 86º.

As a result, the power of the resistor will be 100% real power. There will be no reactive power. In order to have reactive power, you must have reactive elements: inductors or capacitors.


  P = I^2 R = 3.06 mW @ 172º


The formula \$P=I^2R\$ isn't meant for phasor cases. For phasors, you should use

$$P=I^\star IR$$

where \$I^\star\$ is the complex conjugate of \$I\$.

In general, the power consumed by a component with phasor voltage \$V\$ and current \$I\$ is

$$S=VI^\star$$ 

giving complex power \$S\$, with \$S=P+jQ\$ where \$P\$ is real power and \$Q\$ is reactive power.

With this correction you will get a real result, as expected.


  Which figure should I be comparing against a resistor specsheet's power rating?


Since the power used by a resistor is entirely real power, you need to calculate the real power and compare it to the rating.

If the frequency applied is very low, you may need to compare the peak power rather than the average power, since it could be possible for the resistor to burn out at the peak of the waveform.