Burning different isotope combinations of hydrogen and oxygen, how much does the energy vary?
Let me add to Paul's excellent answer with a simple thermodynamic calculation.


Hydrogen oxidation.
$$\ce{H2 + 0.5 O2 -&gt; H2O(g)}$$
This reaction has a standard free energy change of about -229 kJ/mol.  Using $\Delta G^\circ = -RT \ln K$, I get the equilibrium constant for this reaction of $K=1.2\times10^{40}$.
Deuterium isotope exchange between water and hydrogen.

A 1976 paper gives the value for this reaction.

$$\ce{H2O(g) + HD -&gt; HDO(g) + H2}$$
The value given is $K = 3.5$, which by the standards of isotope exchange reactions is very large. Using $\Delta G^\circ = - RT \ln K$, we can calculate that the standard free energy change of this reaction is $\Delta G ^\circ =$ -3.1 kJ/mol.
Add the above reactions. 
$$\ce{HD + 0.5 O2 -&gt; HDO(g)}$$
We can add the free energies of the reactions to get that $\Delta G = $-231 kJ / mol.


Thus, burning $\ce{HD}$ gas provides about 1.4% more free energy than burning $\ce{H2}$ gas. And this is by far the largest difference that we would observe for any common stable isotope.  For nearly any isotopic exchange reaction that involves other isotopes, such as $\ce{^{13}C}$ or $\ce{^{18}O}$, the relevant equilibrium constants will be $0.9 \lt K \lt 1.1$ or so.  Only deuterium vs. protium (i.e. $\ce{D}$ or $\ce{^2H}$ vs $\ce{^1H}$) would often have values outside that range.  Thus for isotopes like $\ce{^{13}C}$ or $\ce{^{18}O}$, the difference will be closer to 0.1%, or, as isotopic geochemists like to say, 1 &permil; (i.e. one per mille).