Einsteins equivalence principle and radiation from accelerating electric charges?

Whoa, great question and I shouldn’t answer it after drinking all day watching college football ...

The answer is no. If you are accelerating WITH the rocket you will not see radiation emitted. 

If you are standing on the ground looking at a charge also on the ground you will not see radiation emitted. 

If you are in a reference frame that is NOT accelerating with the rocket, you will see radiation emitted. 

If you are not in the same non-inertial frame as the charge on the ground you will see radiation emitted.

Burp.

The idea that physics should be the same in all inertial frames was not exactly new to Einstein and the question you ask is essentially what Lorentz was addressing in the 1890's (ie well before Special Relativity) when looking at how to transfer Maxwell's equations between different inertial frames (which they did not naturally do). In order to resolve Maxwell's equations with relative motion between inertial frames, Lorentz determined there was relationship between the frames that boiled down to contraction by a certain factor of gamma (γ), where  γ = √1 - v²/c²

Thus you can see he had already discovered the same formula as used for the time and distance contraction equations of regular bodies, as deduced from consideration of Einstein's "light clock". The difference being that Lorentz did not give his result the physical interpretation that Einstein did.

This seems kind of amazing to a modern scientist - Lorentz had all the information and the mathematical results but couldn't put it together to make a simple interpretation. Interestingly, the same thing happened with Planck and black body radiation. He did all the hard work and noticed that light energy had to be mathematically quantised to avoid the Ultraviolet Catastrophe but it was Einstein who made the step to saying that light was physically delivered by discrete photons.