Darren,

Yes, for a simple resistor with the same Duty Cycle waveform applied to it, it does not matter what the frequency of the waveform is; it will have the same real power dissipated in it.

However, do you think that in CMOS circuits, it takes half a clock cycle for the logic to change states?

I think it is rather the opposite. At a clock edge, there is a 'flurry' of activity as the states change. After this clock edge, no current (except for leakage), flows until the next clock edge. I do not think this 'flurry' of activity changes speed depending on the input clock speed.

Think about these diagrams of the current drawn by an IC, depending on frequency:

A
  ___             ___
  |  |            |  |
__|  |____________|  |___________

B
  ___        ___
  |  |       |  |
__|  |_______|  |______

Looking at these two waveforms, you can see that the hightime in both A & B are the same, as it will be for any frequency the IC supports. What is different is the overall period of the waveform. A is being clocked at a slower rate than B. Thus, the Duty Cycle of the two waveforms is NOT 50%, NOT the same, and is also different for each clock speed. Furthermore, as the duty cycle changes with frequency, the RMS current drawn will increase with the higher clock speed.

Now of course, this has been simplified greatly, but I think (hope?) it shows enough to get the point across.

Regards,
-Dave