When an electron around an atom drops to a lower state, is 100% of the energy converted to a photon?

There is always energy loss due to recoil. Considering a single atom that is in free space, the total momentum is conserved, and the total reaction energy is constrained by the transition energy $E_0$ (ignoring the natural line width due to the energy-time uncertainty), so we get in the centre of mass frame: $$0 = \hbar k + mv $$ $$E_0 = \frac 1 2 mv^2 + \hbar \omega = \frac 1 2 mv^2 + \hbar c k$$ Leading to the equation $$E_0 = \frac {\hbar^2 k^2}{2m} + \hbar c k.$$ So $$k = - \frac{mc}{\hbar} \pm \sqrt{ \frac{m^2c^2}{\hbar^2} + \frac{2mE_0}{\hbar^2} } = \frac{mc}{\hbar} \left( \sqrt{1 + \frac{2E_0}{mc^2}} - 1 \right) \approx \frac{E_0}{\hbar c} - \frac 1 4 \frac{E_0^2}{\hbar mc^3}. $$ This corresponds to an energy correction for the photon of $$\frac{\Delta E}{E_0} = -\frac{E_0}{4 mc^2}.$$ So the relative correction is of the order of the transition energy compared to the rest energy of the recoiling mass.

If the atom interacts with other atoms (and is not in free space) the process becomes complex and energy may be transferred to the interacting objects (e.g. in a crystal lattice) the possible recoil energies are determined by the phonon spectrum, here, at low temperatures, the Mößbauer effect becomes important wherein the recoil momentum is transferred to the entire crystal, leading to a virtually recoil free emission (since the reaction mass is macroscopically large).

Further, there are other ways an electron can loose its energy – e.g. in Auger processes another electron is ejected from the atom, or the energy can be transferred ("coherently") to another atom (exciting an electron there). Further, combination processes, such as emitting a photon and ejecting an electron are allowed if all quantum numbers are conserved.

It depends on the reference system used.

In the center of mass of the excited atom, momentum concervation should assure that the photon and the atom have equal and opposite momenta, so that will diminish the energy taken by the photon. The mass of the nucleus though is so much larger than the ev transitions of the electons that in effect the center of mass is the same as the rest frame of the nucleus.

The energy levels have a width, thus there is a probability of a photon with a smaller or higher frequency than the average of the energy level should appear.

Have a look here for a discussion of broadening of lines due to the motion of the atoms, and also the doppler shift in spectra due to the motion of the source or the detector.

It all depends on the frame of reference.