Do the Earth and I apply the same gravitational force on each other in GR?
For a Newtonian $n$-body system, the weak Newton's 3rd law implies total momentum conservation, but not vice-versa, cf. e.g. this Phys.SE post. However for a 2-body system, which OP asks about, they are equivalent, so OP's question is essentially equivalent to:
How do we see total momentum conservation in full-fledged GR without going to the Newtonian limit?
Answer: That's a great question! Notions, such as, e.g., force, mass, momentum, energy, etc., are notoriously subtle in GR. For a generic spacetime, there is no satisfactory definition of a gravitational stress-energy-momentum tensor, only a pseudotensor.
For an asymptotically flat spacetime, one may define an ADM energy-momentum 4-vector, which plays the role of conserved total energy-momentum associated with the full spacetime (incl. probes).
This answers OP's question in principle, but may not be completely satisfactory: We can easily associate localized energy-momentum to each point-probe$^1$ in the spacetime, but it is less clear how to give an independent definition for the energy-momentum of spacetime minus the probes (other than to declare it to be the difference). I.e. translated back to OP's problem: We don't seem to have an independent definition of Earth's energy-momentum by itself, even if we for simplicity assume that Earth is a black hole with a point-like singularity/matter-distribution.
$^1$ Our notion here of a point-probe is a point particle that can be assigned a localized energy-momentum, but unlike a test particle, it can backreact on the spacetime. The notion of probes is not really essential for the discussion. To emulate OP's 2-body system without using probes, consider instead 2 black holes with point-like singularities/matter-distributions. There doesn't seem to be a well-defined notion of energy-momentum associated to each individual black hole. Their individual energy-momenta are fuzzy/delocalized/ill-defined.
You both fall toward the common center of mass. Because the mass of the Earth is quite a bit larger than yours, the center of mass is very close to the center of the Earth, but rather far away from you. Thus, as you both fall to the common center, the Earth hardly moves while you fly until you hit the ground.
More specifically, you are talking about the two-body solution. Both bodies curve spacetime and move in this curved spacetime. As you justly stated, your contribution is small and for this reason the Earth movement toward you is very small as well.
However, when you interact with the Earth, the momentum you get equals the momentum the Earth gets. And yes, in the classical view, you attract the Earth with the same force as the Earth attracts you. While your gravity is very weak, the mass of the Earth attracted by it is enormous. Therefore the forces work out to be the same, as expected.