Yes. The amount of energy radiated as gravitational waves will depend on the details of the two black holes before the merger. The answers to questions like:
- Was the orbit circular or elliptical?
- Were they spinning?
- Were the spins of the black holes aligned with the orbital plane?
will affect the energy of the gravitational waves. The most import detail in terms of energy radiated is the mass ratio of the two precursor black holes.
The final mass of the system $M_\mathrm{fin}$ is always less than the original total mass ($m_1 + m_2$), since some of the mass energy gets converted to gravitational waves, $M_\mathrm{rad}$.
$$M_\mathrm{fin} + M_\mathrm{rad} = m_1 + m_2 $$
Because of something akin to the second law of thermodynamics the final black hole must be bigger than the biggest original black hole. Basically, you can't radiate so much energy in gravitational waves that a black hole shrinks.
$$M_\mathrm{fin} > m_1 \quad \mathrm{and} \quad M_\mathrm{fin} > m_2$$
We can define the fraction of mass radiated aways as:
$$ e = \frac{M_\mathrm{rad}}{m_1 + m_2} $$
This is sometimes called the efficiency of radiation.
If the black holes have about the same mass (as they did in the LIGO detection), about 5% of the total mass will be radiated away. This is the most efficient possibility.
On the other extreme imagine the case where one black hole is way more massive than the other: maybe 1 million solar masses and 1 solar mass. In order to follow the two rules stated above, $M_\mathrm{fin}$ is less than 1 million and one solar masses and greater than 1 million solar masses. In this case the efficiency would be about $e=10^{-6}$ or 0.0001%. Extreme mass ratios produce the weakest gravitational waves.
You've forgotten an important player in the system: the gravitational field.
Here's a pretty argument that gravitational fields are physically meaningful objects that carry energy: imagine two masses accelerating towards each other from rest, from a great distance away. The rest energy of the system is $E_\text{rest} = (m_1+m_2)c^2$; the kinetic energy is $K\approx\frac12m_1v_1^2 + \frac12m_2v_2^2$, at least while things are nonrelativistic, and only increases as a function of time. We introduce an internal energy $U=-Gm_1m_2/r$ so that we can make statements like "the total energy of the system is constant in time."
Now let's make partitions of our system to see whether we can account for everything. Looking only at the first particle, we see a total energy $E_\text{1} \approx m_1c^2 + \frac12m_1v_1^2$ which starts off positive and grows larger in time. Looking only at our second particle we also see a total energy which starts off positive and grows larger in time. So apparently if we only consider the particles in our system, we can't duplicate our statement that the total energy of the system is a constant in time. We need also to account for the energy tied up in the interaction between the two particles: the gravitational field. In electrodynamics and in general relativity you learn to actually compute how much of this interaction energy $U=-Gm_1m_2/r$ is found in any particular volume of the space around your interacting objects.
When objects emit gravitational radiation without colliding, that radiated energy comes from the gravitational field. Perhaps better, gravitational radiation is a redistribution of the energy stored in the gravitational field: energy is removed from the field near the interacting particles, leaving them more tightly bound to one another, and appears at large distances from them, where it can do things like move interferometer mirrors.
When you have nonrelativistic objects collide, you have conversion of gravitational energy into other forms of internal energy, like heat; this is why asteroid impacts can melt things. Eventually the heat gets radiated away, too.
A black hole is an object whose total energy is stored in the gravitational field --- we talk about a black hole's mass as a shorthand for how much of this gravitational energy there is.
Best Answer
Suppose you have two black holes of the same mass $M$ and $m = GM/c^2$. The radius of each black hole is then $r = 2m$, and the horizon area is $A = 4\pi r^2$ $ = 16\pi m^2$. Two constraints are imposed. The first is that the type-D solutions have timelike Killing vectors, which are isometries that conserve mass-energy, and with the merger the gravitational radiation is in an asymptotically flat region where we can again localize mass-energy. So the initial mass $2M$ is the total energy. The entropy of the two black holes is a measure of the information they contain and that too is constant. So the horizon area of the resulting black hole is the sum of the two horizon areas, $A_f = 2A$ $ = 32\pi m^2$, that has $\sqrt{2}M$ the mass of the two initial black holes. Now with mass-energy conservation $$ E_t = 2M = \sqrt{2}M + E_{g-wave} $$ and the mass-energy of the gravitational radiation is $.59M$. That is a lot of mass-energy!
This is the upper bound for the generation of gravitational radiation from mass. The assumption here is that the total entropy of the two black holes equals the entropy of the final black hole. Physically this happens if all the curvature exterior to the merging black holes does not result in mass-energy falling into the final black hole. There would be back scatter of gravitational radiation, much as one has to be concerned about the near field EM wave near an antenna that can couple back on it. The final entropy of the merged black hole will in fact be larger, but of course not larger than the mass-squared determined area of the two black holes. This means $1.41m~\le~m_{tot}~\le~2m$.
To estimate this requires numerical methods. Larry Smarr pioneered a lot of this. So far estimates run around $5\%$ of the total mass of the black holes is converted to gravity waves. in this LIGO paper two black holes of mass $39M_{sol}$ and $32M_{sol}$ is computed to have coalesced into a final black hole of $68M_{sol}$, which radiated $3M_{sol}$ is gravitational radiation and accounts for $4.2\%$ of the initial mass. This is about in line with most numerical studies. Consequently a lot of the spacetime curvature generated by these mergers falls back into the final black hole. In terms of area the initial horizon area is $4066M^2_{sol}$ and the final horizon area is $4624M^2_{sol}$, which is an additional area $558M^2_{sol}$ of horizon area with $S~=~k A/4L_p^2$ as the entropy.