Let us first discuss why do usual bodies absorb electromagnetic waves so well. A normal body is made of protons and electrons, which have specific charges of opposite sign and vastly different magnitude. As a result, every atom will be strained by the electromagnetic wave even if its size is completely negligible with respect to the wavelength because they feel opposing forces at any point. Zooming out, atoms and/or molecules will have a magnetic dipole moment, which can also accelerate it as a whole. If the body is made out of metal, free electrons react to the wave immediately, bump into the protons in the metal lattice, and cause heating. All these phenomena come about because the body is made up of elements that react differently to the passing electromagnetic wave, and this is also the reason why usual matter is such a good absorber of EM radiation.
Now let us take a look at gravitational waves. Here the situation is very different since every piece of matter in the Universe has the same "specific charge" with respect to gravity. (This is called the equivalence principle and we know it from our everyday as "gravity accelerates everything equally".) When the gravitational wave then passes through a structure much smaller than its wavelength, the entire structure is moved in sync. It is like a small boat on a tsunami wave far from shore: nothing really happens.
So if any absorption comes about, it will be from the fact that the body is slightly deformed on scales comparable to the wavelength. This causes internal friction and ultimately also some sort of GW back-reaction from the body. 104
We can approximate the back-reaction by assuming that the energy dissipated by the internal friction $\Delta E$ must be compensated by a change to the momentum of the body $\Delta p = \Delta E/c$ as you state in your question. So yes, some momentum transfer can occur for bodies of comparable or larger size than the wavelength and large internal friction.
However, even a pretty powerful wave will cause modest strains far from the source. It can be shown that a GW source of mass $M$ will cause a strain at most $\lesssim GM/(c^2 R)$ where $R$ is the distance from the source (the upper bound is satisfied only for equal-mass binaries at merger). For example, if you wanted to cause a strain of $10^{-2}$ at a distance of $100$ meters, you would need a generating mass of the order of magnitude of $\sim 100$ Earths (in a black hole binary near to merger). It is good to note, though, that no object of size smaller than $GM/c^2$ would ever be strongly affected by the wave since that is the lower bound for the wavelength.
About the "GW pressure" overpowering the gravitational attraction. It is good to again remember that gravity has only one sign of "charge". In electromagnetism, we can put a positive and negative charge together, make them oscillate, and the resulting oscillating dipole has no net attraction or repulsion for any single particle. This cannot be done in gravity, every source of radiation in gravity comes with an attraction that satisfies certain inequalities with respect to the radiation field. For instance, if you want to collimate your wave, you cannot do it without putting masses next to each other, there is no such thing as a "GW reflector". At this point I offer a conjecture anyone is free to explore: I believe that if you go into the question of how much momentum can be absorbed by the internal friction and put fundamental physicality bounds on the friction coefficients, the "GW radiation pressure" will never be stronger than the attraction apart from a special limiting configuration at the moment of the merger. 70