Explicit "soft" breakdown of symmetry (i.e., the operator that breaks the symmetry has mass dimension $<4$) leads to a pseudo Nambu-Goldstone boson (pNGB) with "small" mass. How does one show that, in a simple case, the radiative corrections do not drive its mass to infinity or to a very large value?
In other words, if we broke the symmetry explicitly by an operator having mass dimension $\geq 4$, how does the would-be pNGB becomes heavy?
When the symmetry is broken explicitly by a relevant operator, its contribution is negligible at high energies. Thus, if $f$ is the scale of the breaking operator, at energies $\Lambda\gg f$ the symmetry is restored. If there is a spontaneous breaking of the symmetry, there will be massless Nambu-Goldstone bosons, as usual.
This is a high energy description, so these (pseudo-)Nambu-Goldstone bosons are not really massless. What happens is that their masses $m$ are so small that they can't be noticed at high energies ($\Lambda\gg m\sim f$). Therefore, the mass is protected by the appearance of the symmetry at high energies.
When the operator is irrelevant, the symmetry is not restored at high energies, a cut-off scale is introduced and the masses should in principle (and maybe a little bit naively) receive corrections of the order of this scale, which should be at least as large as the range of energies the theory is able to describe.
For marginal operators, a general answer can't be given, as the renormalization group behavior depends on the particular case.
Say that one breaks a continuous symmetry with an operator parametrized by a dimensionless parameter $h$. Since the mass of the Goldstone modes is zero for $h=0$, we expect the mass to be generically of the order of $h$ (or to some power of $h$), so in principle the mass can be as small as one want, as long as $h$ is small enough (a more explicit criterion would depend on the model).
The other question of the OP, concerning the effect of an irrelevant operator is quite interesting, and is not discussed much in the literature. However, there has been a renewed interest in this question, and a nice reference is arXiv:1508.07852. I'll give here a brief summary of the conclusion of the paper.
In the case of an irrelevant (for the gaussian fixed point) symmetry breaking field, one seems to arrive at a contradiction. Since $h$ breaks the symmetry, we expect a finite mass for the would-be Goldstone modes, but the fact that $h$ is irrelevant would naively tell us that we can put $h=0$ directly from the beginning, which would lead to truly massless Goldstone modes.
The contradiction is lifted by the fact that $h$ is in fact a dangerously irrelevant perturbation, that is, one that scales to zero at the gaussian fixed point, but that we cannot put directly to zero in the action. More precisely, $h$ is indeed irrelevant at the gaussian fixed point, but strongly relevant at the infrared (ordered phase) fixed point.
A schematic description of the RG flow is the following (assuming $h$ is small, and that the interaction is also small, so that the gaussian fixed point describes well the UV). During the first part of the flow, the system believes it is gaussian. $h$ flows to zero, but the interaction is relevant. At some point, the system realizes it is not gaussian, and flows towards the Goldstone fixed point. Because $h$ is now very small, the system can spend quite some time believing it has a continuous symmetry, and in the intermediate regime, will behave as if it had true Goldstone modes. Finally, deep in the IR, the system realizes that there is an explicit breaking of the symmetry, which generates a (very) small mass for the Goldstone modes.
The two main consequences of this mechanism is that the mass will be generically (i.e. without fine tuning) very small compared to the UV energy scale, and that the critical exponents (in the case of 2nd order phase transitions) will be different on both side of the transition (since in the disordered phase, the breaking field does not play any role).
