As you say, loop corrections change the value of the quartic coupling $\lambda$ is modified by loop corrections. If the renormalization group changed its value so $\lambda \gg 1$, the perturbative interpretation of the theory would break apart. Note that the condition $\lambda < 1$ (or $\lambda/(4\pi) < 1$) is only a hint of the perturbativity, the real condition is that the two-loop correction (order $\lambda^2$) must be neglible with respect to the one-loop correction.
Note also that the real danger isn't non-perturbativity, but the existence of a <PERSON> pole, i.e., that the coupling constant goes to infinity at some finite energy. Of course, non-perturbativity is usually a warning of a <PERSON> pole. But you could encounter "new physics" at some energy that modifies your beta function and makes the theory perturbative again. 976
In that case, non-perturbativity only conveys our failure on doing the maths (well, the necessity of using other approaches such as lattice field theory).
But... in the SM that doesn't happen [the question asks specifically for SM, even though OP is concerned about other models in comments below]. The one-loop Renormalization Group equation is $$16\pi^2 \frac{d \lambda}{d(\ln \mu)} = 12 \lambda^2 + 12 y_t^2\lambda -12 y_t^4 $$
where $y_t$ is the Higgs-top Yukawa coupling, and the rest of Yukawa couplings and gauge group couplings have been neglected. The renormalization group equation is a first-order differential equation. So, after plugging the value of $\lambda$ at one energy scale, you can predict its value at any other energy. Usually we take that energy as the electroweak scale $\mu_{EW} \sim 100 \mathrm{GeV} \sim v$, where the spontaneous symmetry breaking occurs, and use the value you quote $\lambda(\mu_{EW}) = m_h^2/2v^2$.
As the energy scale $\mu$ grows, $\lambda(\mu)$ gets lower, and around $\mu\sim 10^{10}\mathrm{GeV}$, $\lambda(\mu)<0$.
What does this mean? At large energies ($\mu \gg m_h$), $\lambda$ determines completely the Higgs potential, and if $\lambda<0$, then the potential is metastable, it can decay to a lower energy state of the Higgs field.
Both figures are taken from <PERSON> et al.: Higgs mass and vacuum stability in the Standard Model at NNLO arXiv:1205.6497 [hep-ph] 976