Since he doesn't manage to make it back to the runway from his 2000 ft engine flameout run I was wondering if not deploying the RAT would make any difference due to reduced drag?
No, because it deploys automatically. Note that in the video in the first attempt the step is not done because there is less time. It is probably included in the checklist just in case the automation didn't do it's job.
Nevertheless, let's try to estimate the drag. All I could quickly find about the RAT is that the attached electric generator provides 5 kW, but that should be enough. We need to take into account the hydraulic pump and power losses, but the generator wouldn't be insignificant part of the load, so we could guess that the turbine does not have more than 50 kW of drag.
Now potential energy is $E_p = mhg$, differentiating by time we get $P = v_{vert}mg$ and solving for vertical speed $v_{vert} = \frac{P}{mg}$. Taking some “median” weight of 60 t (A320 has MZFW 62.5 t and MLW 66 t) and substituting we get:
$$v_{vert} = \frac{50\ \mathrm{kW}}{6\cdot10^4\ \mathrm{kg}\cdot 9.8\ \mathrm{ms}^{-2}} \approx 0.085\ \mathrm{ms}^{-1} \approx 17\ \mathrm{ft/min}$$
So the deployed RAT increases the rate of descent by less than 20 ft/min.
Now the video mentions the best glide speed would be 205 knots (I suppose it is for the 1+F configuration; for clean it sounds too low, though I don't have a reference for this). Since the L/D ratio is around 18, that would give the total descent rate about:
$$v_{vert} = \frac{205 \mathrm{knot}}{18} \approx 11 \mathrm{knot} \approx 1150 \mathrm{ft/min}$$
Except, well, 18 is clean L/D. The L/D at 1+F is a bit lower. I don't have reference for that either, but something like 10–12 sounds right, so:
$$v_{vert} = \frac{205 \mathrm{knot}}{12} \approx 17 \mathrm{knot} \approx 1730 \mathrm{ft/min}$$.
That is about 100 times more. So the RAT might increase drag by 1%. It was very rough estimate, so it may be 2% or 0.5%, but it definitely isn't that significant.
Would the on-board batteries and windmilling engines be able to provide sufficient power for control surfaces?
The training manual mentions that
If engine wind milling is sufficient, additional hydraulic power may be recovered.
without saying at what speeds that would be, but generally the speed for wind milling engine restart are significantly higher that the best glide speed. I'd say the necessary speed would be more than 250 knots.
Also, are there real life incidents of a 2000 ft (or lower) dual engine failure that made a safe return to the runway? Sully Sullenberger's A320 had reached approx. 3,000 ft, yet he was forced to ditch rather than return.
I can't find any other incidents involving dual engine failure on A320-family aircraft besides US Airways 1549. And I am not aware of any incidents on any other aircraft that would fit that criteria either.
It should however be noted, that flight 1549 likely had enough altitude to return to the runway—if the manoeuvre was executed perfectly. A small mistake and they'd crash into populated area since the airport is in the middle of a city. The Sullenberger's great airmanship was in that he chose the worse option that he was certain he can make over the better option where he was not.
Does the energy management differ drastically from type to type? i.e. Among dual engined aircraft would some be easier to land from a 2000 ft dual engine failure than others?
Principally it's the same for any type. But since turn radius increases drastically with speed, slower aircraft can turn faster and thus could return from lower altitude.
Here's a related question that discusses the manoeuvre; and it's dangers: Is it even remotely feasible to turnback a single engine aircraft with an engine failure?
