If by "fully classical", you mean a theory defined by only tree level Feynman rules, then yes. You will drop all higher powers of $G_N$ in the bulk. And yet, you will not be able to drop any "loops" when trying to match these calculations with the boundary CFT. Instead what you will drop are terms of order $1/N^2$ in the large $N$ expansion of the gauge theory. The weak coupling expansion will not truncate because this requires $\lambda = g^2_{\text{YM}} N$ to be small. The 't Hooft limit you mention just sets it to be fixed and to recover supergravity, you need something stronger than just the 't Hooft limit. Specifically, $\lambda$ needs to approach infinity but just not as quickly as $N$.
Of course, when I say that additional powers of $G_N$ make the bulk theory more quantum while additional powers of $\lambda$ make the boundary theory more quantum, this is just because "interaction Lagrangians" are proportional to these coupling constants. But perturbation theory around a free theory doesn't just bring down powers of $\int \mathcal{L}_{\text{int}}$... rather $\frac{i}{\hbar} \int \mathcal{L}_{\text{int}}$. So it is common to call tree level "classical".
There are a couple caveats to "tree level = classical". For instance, some people would say that observables like cross sections and decay rates are quantum even when you only compute them at tree level because they are defined using a Hilbert space. So in this sense "classical graviton S-matrix" might be an objectionable term because the very existence of gravitons is a sign of something quantum. The other caveat is that truncating to tree level is only valid when the loop counting parameter is smaller than every other parameter. So certain problems actually cannot be treated classically even when we are okay with taking $1/N$ and therefore $G_N$ to be small. An example is a graviton state evolving into a black hole. The operators dual to black holes in $\mathcal{N} = 4$ SYM involve $O(N^2)$ fields so the diagrams beyond the planar limit are suppressed as we expect but there are now so many of them that they all need to be kept (perhaps modulo some highly non-trivial effects). This is the logic Witten used to explain why baryons are harder to deal with than mesons.
I think the second caveat is more serious. If you take $\alpha' / L^2 \to 0$ then the bulk will be governed by a supergravity EFT. All stringy modes except the lowest (which look like gravitons) will be gone. And scattering amplitudes for these gravitons will also be unaffected by stringy effective action terms like $R^4$. But saying this EFT is classical when $N$ is large enough needs to be taken with a grain of salt because some processes will still be quantum. This probably shouldn't surprise us in view of the information paradox.
