OP asks a good question about the interplay between the classical and the quantum theory in Goldstone's theorem. In this answer we review Goldstone's theorem following Refs. 1 & 2.
The typical starting point is an (infinitesimal) zero-mode quasi-symmetry transformation
$$\delta\phi^{\alpha}(x)~=~\epsilon c^{\alpha}{}_{\beta}\overline{\phi^{\beta}}, \qquad \delta\widetilde{\phi}^{\alpha}(k)~=~(2\pi)^d\delta^d(k)\epsilon c^{\alpha}{}_{\beta}\overline{\phi^{\beta}},\tag{1} $$
of the classical action $$S_V[\phi]~:=~\int_V\! d^dx~{\cal L}$$
(where we assume that the Lagrangian density ${\cal L}(\phi(x),\partial\phi(x))$ has no explicit $x$-dependence)$^1$:
$$\begin{align}0~\sim~\delta S_V[\phi]~\sim~&
\int_V\! d^dx~\frac{\delta S[\phi]}{\delta\phi^{\alpha}(x)}\delta\phi^{\alpha}(x)\cr
~\stackrel{V=\mathbb{R}^d}{=}&~\int\! \frac{d^dk}{(2\pi)^d}~\frac{\delta S[\phi]}{\delta\widetilde{\phi}^{\alpha}(k)}\delta\widetilde{\phi}^{\alpha}(k).\end{align}\tag{2}$$
Here we have introduced the spacetime-averaged quantity
$$ \overline{f(\phi)}~:=~\frac{1}{|V|}\int_V\! d^dx~f(\phi(x))\tag{3} $$
of a spacetime region $V\subseteq\mathbb{R}^d$.
Now the main point is that the 1PI quantum effective action$^2$ $\Gamma_V[\phi_{\rm cl}]$ inherits$^3$ the zero-mode quasi-symmetry (2) of the classical action
$$\begin{align}0~\stackrel{(2)}{\sim}~\delta \Gamma_V[\phi_{\rm cl}]~\sim~&
\int_V\! d^dx~\frac{\delta \Gamma_V[\phi_{\rm cl}]}{\delta\phi_{\rm cl}^{\alpha}(x)}\delta\phi_{\rm cl}^{\alpha}(x)\cr
~\stackrel{V=\mathbb{R}^d}{=}&~\int\! \frac{d^dk}{(2\pi)^d}~\frac{\delta \Gamma_V[\phi_{\rm cl}]}{\delta\widetilde{\phi}_{\rm cl}^{\alpha}(k)}\delta\widetilde{\phi}_{\rm cl}^{\alpha}(k).\end{align}\tag{4}$$
Functional differentiation then yields$^4$
$$0~\stackrel{(4)}{=}~
\int_V\! d^dx~\frac{\delta^2 \Gamma_V[\phi_{\rm cl}]}{\delta\phi_{\rm cl}^{\beta}(x^{\prime})\delta\phi_{\rm cl}^{\alpha}(x)}c^{\alpha}{}_{\gamma}\underbrace{\overline{\phi_{\rm cl}^{\gamma}}}_{=\overline{\langle\phi^{\gamma}\rangle_J}}+\underbrace{\frac{\delta \Gamma_V[\phi_{\rm cl}]}{\delta\phi_{\rm cl}^{\alpha}(x^{\prime})}}_{=-J_{\alpha}(x^{\prime})\approx 0}c^{\alpha}{}_{\beta}\frac{1}{|V|}
.\tag{5}$$
The Fourier transform reads:
$$\frac{\delta^2 \Gamma_V[\phi_{\rm cl}]}{\delta\widetilde{\phi}_{\rm cl}^{\beta}(k^{\prime})\delta\widetilde{\phi}_{\rm cl}^{\alpha}(0)}\underbrace{\overline{c^{\alpha}{}_{\gamma}\langle\phi^{\gamma}\rangle_{J=0}}}_{\neq 0}~\stackrel{(5)}{\approx}~0.\tag{6}$$
Let us now rename one of the (possible several) Goldstone fields as $\pi:=c^{\alpha}{}_{\beta}\phi^{\beta}$ with non-zero VEV $$\overline{\langle \pi\rangle_{J=0}}~\neq~ 0.\tag{7}$$
It follows from eqs. (6) & (7) that the Hessian
$$\left.\frac{\delta^2\Gamma[\phi_{\rm cl}]}{\delta \widetilde{\pi}_{\rm cl}(k_1)\delta\widetilde{\pi}_{\rm cl}(k_2)}\right|_{k_1=0=k_2}~\stackrel{(6)+(7)}{=}~0\tag{8}$$
has a zero, or its inverse$^5$, the Fourier-transformed connected propagator/2-point function
$$\langle\widetilde{\pi}(k_1)\widetilde{\pi}(k_2)\rangle^c~\stackrel{(8)}{\propto}~\frac{1}{k_1^2}\delta^4(k_1\!+\!k_2)\tag{9}$$
has a pole,
i.e. $\pi$ is massless, thereby proving the Goldstone theorem. $\Box$
References:
M.E. Peskin & D.V. Schroeder, An Intro to QFT, 1995; Section 11.1 p. 351-352 + Section 11.6 p. 388.
S. Weinberg, Quantum Theory of Fields, Vol. 2, 1995; Section 19.2 p. 167-168.
S. Weinberg, Quantum Theory of Fields, Vol. 2, 1996; Section 16.4 p. 77 + Section 17.2 p. 84.
$^1$ Notation: The $\sim$ symbol means equality modulo boundary terms. The $\approx$ symbol means equality modulo eqs. of motion.
$^2$ At tree-level/zeroth order in $\hbar$, the quantum effective action $\Gamma_V[\phi_{\rm cl}]$ is the classical action $S_V[\phi_{\rm cl}]$, cf. e.g. eq. (12) in my Phys.SE answer here.
Much of the construction can equivalently be reformulated in terms of the quantum effective potential density
$$ {\cal V}_{\rm eff}(\overline{\phi_{\rm cl}})~:=~-\frac{1}{|V|}\Gamma_V[\phi_{\rm cl}\!=\!\overline{\phi_{\rm cl}}]. \tag{10} $$
$^3$ More generally, the generating functional $W_{c,V}[J]$ of connected diagrams and the 1PI quantum effective action $\Gamma[\phi_{\rm cl}]$ inherits affine quasi-symmetries of the classical action $S_V[\phi]$ and the path integral measure, cf. e.g. Ref. 3. In the present case the sources transform as
$$ \delta J_{\alpha}(x)~=~-\epsilon \overline{J_{\beta}} c^{\beta}{}_{\alpha}, \qquad \delta\widetilde{J}_{\alpha}(k)~=~-(2\pi)^d\delta^d(k)\epsilon \overline{J_{\beta}} c^{\beta}{}_{\alpha},\tag{11} $$
and the classical fields as
$$\delta\phi_{\rm cl}^{\alpha}(x)~=~\epsilon c^{\alpha}{}_{\beta}\overline{\phi_{\rm cl}^{\beta}}, \qquad \delta\widetilde{\phi}_{\rm cl}^{\alpha}(k)~=~(2\pi)^d\delta^d(k)\epsilon c^{\alpha}{}_{\beta}\overline{\phi_{\rm cl}^{\beta}}.\tag{12} $$
$^4$ Recall that $\phi_{\rm cl}^{\alpha}=\langle \phi^{\alpha}\rangle_J$, cf. e.g. eq. (4) in my Phys.SE answer here.
$^5$ See e.g. eq. (8) in my Phys.SE answer here.
