In the Schwarzschild black hole, the Killing vector "time translation" $k^a$, so that the following quantity is conserved along a geodesic:
$$E = -g_{ab}k^au^b = (1 - \frac{2GM}{r})\frac{dt}{d\tau}.$$
Which is interpreted as the total energy per unit mass measured by a static observer. However, a body orbiting a black hole will radiate part of its energy in gravitational waves.
What is the phyisical interpretation? Would this mean that energy is not conserved?
The conservation of $\vec{k}\cdot\vec{u}$ only holds in the test particle limit. That is, it considers the metric to be unaffected by the motion of the particle. In this limit, there are no gravitational waves, since the metric has no time-varying quadrupole.
If you want to see gravitational waves, you need to allow the metric to evolve dynamically, considering the motion of the particle (whose mass must then enter into the problem). In this case that same $\vec{k}$ is no longer a Killing vector.
Since the spacetime is still asymptotically flat you can make some sense of a global energy quantity that is conserved, but it will be a sum of the particle's kinetic energy, the gravitational potential energy of the system, and the energy in gravitational waves.
