I understand that a general interpretation of the $1/r^2$ interactions is that virtual particles are exchanged [...] General relativity does not suppose that zero-mass particles exchanged.
You don't need quantum field theory for this. In a purely classical field theory, we have field lines, and the field lines from a spherically symmetric source should radiate outward along straight lines. In a frame where the source is at rest, we expect by symmetry that the field lines are uniformly distributed in all directions. The strength of the field is proportional to the density of the lines, which falls off like $1/r^2$ in a three-dimensional space.
This whole description is complicated by the polarization of the field. Gravitational fields have complicated polarization modes. Nevertheless, the $1/r^2$ result is unaffected.
Finally, we have an issue unique to GR, which is that the field is the metric, and this means that the field itself affects the measuring tools that we use to measure things like $r$, the field, and the area of a surface through which we're counting the number of field lines that penetrate. These are all strong-field issues, so for large $r$, they don't affect the $1/r^2$ argument.
Is it a postulate?
No. In the standard formulation of GR, the main postulate is the Einstein field equations. From it, we can prove Birkhoff's theorem, which says that the Schwarzschild metric is the external field of a static, spherically symmetric source. The weak-field limit of the Schwarzschild metric corresponds to a $1/r^2$ field.