There are a couple of ways to look at this question. The easiest is in first principals.
Therefore imagine that the concord (flying at Mach 2 at FL 600) can suddenly and without energy loss pull up, such that it flies for example 10 degrees upwards. Also we assume the following:
- The engine flames out immediately, therefore no engine thrust.
- We neglect air resistance
How much higher would it go? The formula to calculate this is suprisingly simple:
$$ \Delta x = 1/2 \cdot \frac{(V_I \cdot sin(\gamma))^2}{g} $$
with $V_I$ being the inital velocity (590 m/s which is Mach 2), $\gamma$ being the climb angle and $g$ being the gravity of $9.81 \frac{m}{s^2}$.
I have calculated it for a couple of angles:
- $\gamma = 10°$: 535m
- $\gamma = 30°$: 4436m
- $\gamma = 45°$: 8871m
- $\gamma = 60°$: 13307m
- $\gamma = 90°$: 17742m
That gives us a first guess, but the question remains: Could it pull up at this high altitude? The answer is: Yes. Consider the formula to compute the dynamic pressure:
$$ q = 1/2 \cdot \rho \cdot V^2 $$
The concorde takes-off at around $113 \frac{m}{s}$ at an air density of around $1.225 \frac{kg}{m^3}$. The air density at FL 600 is around 1/10th of the air density at sea level. However the crusing speed is roughly 5.2 times faster, which yields 2.7 times higher dynamic pressure at that altitude. Thus, the rudders definitly work at that altitude.
The real limiting factor here are the structural limits when you try to pull up at Mach 2. For example if you try to pull up your aircraft at Mach 2 with a constant acceleration of 1 G, you will follow an circle with a radius of ~50km. Even with 3 G acceleration, you will still follow an (upward) circle with a radius of roughly 16km. My guess is that the Concord can safely handle around 3 Gs (For example to be able to handle gusts at altitude). Additionally, you will loose a lot of your energy while pulling up, due to additional drag (remember, because of the increase angle of attack, the aircraft has more drag) and of course you are getting slower as you trade velocity versus height. Thus as some point you will not be able to pull-up any further because you lost too much speed. I tried to estimate how much higher you would have to fly in order to go down to stall speed1, and my result is 11.2 km of additional height. That would mean that you could reach a $~70°$ climb. However I neglected air-drag etc. and the fact that you will definitly not pull 3Gs at stall speed.
To get a definit answer, one would have to perform some computer simulations. Perhaps one of the popular flight simulations can help. But I would guesstimate, that at FL 600 going at Mach 2, you can probably (safely) pull the airplane up into a climb of around $\gamma = 30°$, which would yield around ~4.4km of additional height.
P.S: However the assumption that the engine will flame out at higher altitudes is probably true. My guess would be that the engine was designed such that it can not substantially exceed FL 600. But that is speculation.
1 I simply assumed that the stall speed is at the same dynamic pressure as at take-off condition. This yields a stall speed of $357 \frac{m}{s}$