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
Let's start by noting an alpha particle (aka a Helium-4 nucleus, aka 2 protons and 2 neutrons) has a mass of less than the sum of 2 free protons and 2 free neutrons. The reason is that there is some binding energy that is released when the alpha particle is formed. Using $E=mc^2$, some of the mass of the original particles is released as energy when the bound alpha state forms. In equations, \begin{equation} 2 m_p + 2m_n = m_{\alpha} + E_B \end{equation} where $m_p$ is the proton mass, $m_n$ is the neutron mass, $m_{\alpha}$ is the mass of an alpha particle, and $E_B$ is the binding energy.
The same logic holds General Relativity. There are some technical points about how to define energy in GR, but for asymptotically spacetimes one can define the ADM mass, which is a measure of the energy of the entire spacetime and is conserved, and the Bondi mass, which can be used to measure the energy lost due to the emission of gravitational waves. The net result is that
\begin{equation} m_1 + m_2 = M_f + E_{\rm GW} \end{equation} where $m_1, m_2$ are the initial masses of the component black holes of the binary, $M_f$ is the final mass of the remnant black hole, and $E_{\rm GW}$ is the energy emitted in gravitational waves.
$E_{\rm GW}$ can be calculated in post-Newtonian theory (at least the contribution from the inspiral), and depends on the initial masses (if we are dealing with black holes, it is a homogenous function of $m_1$ and $m_2$). (More accurate estimates can be obtained with numerical relativity). For GW150914 (the first binary black hole system to be detected with gravitational waves), the energy emitted was about $3$ solar masses. Gravitational wave events are quite extraordinary; even though they are very difficult to detect, for a brief shining moment near the merger they outshine the entire electromagnetic spectrum with gravitational-wave power.