There are many fundamental concepts that both you and Trimok have misunderstood.
First of all, you can completely ignore the mass energy. The kinetic energy in no way "compensates" for the loss in mass energy because no mass energy is lost in the example that you gave. Sure, the mass of the rocket decreased as it ejected propellant to move forward, but that mass hasn't disappeared. The propellant is still out there, floating around in space. You can't just ignore its mass energy because it's not in the rocket anymore, it still exists. In other words your total $E_m$ term is always the same throughout this situation, at no point is matter converted into another form of energy, like kinetic energy as you seem to suggest.
Second, this is conceptually wrong :
The gravitational attraction between me and the earth will significantly decrease and after a certain distance it will cease to act as the inertia of my rocket would be bigger than the gravitational force between my rocket and the earth.
You can't compare the inertia of a system with a force (they're not even measured in the same unit). One can't be "greater" than another. Regardless of the mass of an object, however small a force you apply on it it will still have an impact. There's no cutoff. This doesn't matter much because indeed, you can get to a point where this is small enough that you consider yourself to be in a situation where the potentially energy is now maximal (you seem to think that because this is 0 the potential energy has now disappeared, but this is not the case, it is sometimes chosen to be 0 at infinity but keep in mind that this is still an increase because in such conventions this potential energy is negative at lift off). Potential energy increases as you get further from the earth. This is because, intuitively enough, now that you are further from the earth you now have the potential to gain more kinetic energy by falling back towards the earth through a longer distance, gaining more speed in the process.
Finally, let's work out the actual energy distribution throughout all this. Like I said we can ignore mass energy because the total mass is conserved. Instead what we actually need to consider is :
$E_k$ : the kinetic energy of the rocket.
$E_c$ : the chemical potential energy stored in the propellant. It is this energy that will propel the rocket, not mass energy as you seem to think.
$E_g$ : the gravitational potential energy.
In the beginning, the rocket is sitting on the surface of the earth, and $E_c$ is maximal, $E_g$ is minimal, and $E_k$ is just $0$. As you start burning propellant, you release the energy stored in your propellant and $E_c$ begins to decrease as $E_k$ increases. To completely compute $E_k$ you actually have to take into account both the kinetic energy of the rocket and the kinetic energy of the ejected propellant. If you do, you will notice that this is actually not enough to make up for the loss in $E_c$. Indeed, the difference between the two corresponds to the gain in $E_g$ as the rocket goes further and further away from the earth, and indeed we will always have conservation of $E_k + E_c + E_g$.
These kind of effects apply to any kind of mass or energy. A massive particle moving at below the speed of light will carry a "subluminal gravitational wave". Dipolar radiation will carry a "dipolar gravitational wave" etc.
In principle, you can assign the label "gravitational wave" to any kind of gravitational field propagating along with some matter-energy object. But this is not usually done. A more purist viewpoint would be to call the free gravitational waves as the only gravitational waves. I.e., if it is a vacuum excitation of the space-time itself, it is a gravitational wave, if it is actually the gravity associated with some object, it is not. Such gravitational waves radiated from isolated sources can indeed be only quadrupole.
However, this line is harder to draw in the case of null (moving at speed of light) matter. In that case, you often do not know which part is the "free" wave and the one "carried along" because they would both move at the speed of light along the matter/radiation. And this is also a fundamental issue; there is no truly fundamental difference between the "space-time at rest" and the "vibrating space-time" through which the wave is passing. We are able to speak of the difference between a "wave" and a "non-wave" only thanks to a predefined meaning of a "still background". But when you are for example in the thin shell of gamma radiation travelling away from a source, the "correct background" is basically impossible to define.
Hence, people often talk about waves dragged with null matter simply as gravitational waves and do not make a distinction between the "free" and "dragged along" part (Griffiths & Podolský have a few chapters with examples). In the paper you link, the distinction could be made because one could show that there is no extra freedom in the polarization or strength of the wave. Simply put, the gravitational wave has no free properties and it is fully determined by the shell of gamma radiation (up to non-physical gauge transformations). So we could either call the mentioned excitation in the metric a "monopole wave" or a "dragged-along gravitational field", it is really just a question of a finer naming convention.
Best Answer
When your going up a hill at a high rate of speed chances are that you started on a horizontal surface going at a high speed and made your way up the hill at a quicker rate so while you are actually going up the hill you are useing the momentum that you got from the above mentioned horizontal surface so it definitely is easier.