Quantum system can emit only photons with energy equal (within the uncertainty) to the difference between two energy states.
Even if the atom is in a superposition of energy states
\left|\Psi\right> = C_0 \left|0\right> + C_1 \left|1\right> + C_2 \left|2\right> + \ldots \qquad (1)
with average energy somewhere between the levels, it can emit only certain set of photons: $E_1 - E_0$, $E_2 - E_0$, $E_2 - E_1$ etc.
Emission of a photon is an act of measurement since the energy of the emitted particle contains information about the atom. If the energy of the photon is $E_2 - E_1$ then the energy of the electron in the atom is $E_1$ - the energy of the final state of the transition. The next photon emitted by this atom will have energy equal to $E_1 - E_0$ for sure.
If one observe photons emitted by an ensemble of atoms in state (1) he will see $E_1 - E_0$ photons with probability $\left|C_1\right|^2$, any of $E_2 - E_0$ and $E_2 - E_1$ with probability $\left|C_2\right|^2$ and so on.
The total energy emitted by the system while it is coming to ground state is equal to average energy of state (1) multiplied by the number of atoms in the ensemble.
Energy conservation is not violated.
The same is true for mixed states for which the probability of certain photon is determined by the density matrix of the system.