Replacing D cell batteries with power adaptor
And unless my math has failed me, the bursts are 1000J so I would need
  about a 55F capacitor, which is out of the question.


Your math has failed you. :-)

You mention that the duty cycle is 2% at 1/3 Hz. The period is therefore 3 seconds, and only during 2% of those 3 seconds does your device deliver 10 A. 2% of 3 seconds is 0.06 seconds.

While your device may consume 60 W while delivering 10 A, it does so for only 0.06 seconds, then "rests" for 2.94 seconds. This means that the energy delivered per pulse is only 3.6 J, not 1000 J:

$$60\text{ W}\times0.06\text{ s} = 3.6\text{ W}\cdot\text{s} = 3.6\text{ J}$$

This amount of energy can definitely be stored in a relatively non-exotic capacitor.

You may not even require a huge power supply. If the device was designed by someone competent, it would already have its own specialized power supply to deliver those 10 A pulses 2% of the time, so while the device can deliver 10 A pulses, it doesn't consume anywhere near 10 A.