$\def\ii{{\rm i}}
\def\dd{{\rm d}}
\def\ee{{\rm e}}
\def\Tr{{\rm Tr}}
$As far as I know and would expect, the replacement of a quantum heat reservoir with a noisy classical field cannot be rigorously justified in general. However, for the simple problem of pure dephasing posed here, there is indeed a correspondence between the quantum and classical noise models. The quantum model has Hamiltonian $H = H_S + H_B + H_{SB}$, where the system Hamiltonian is $H_S = (\omega_0/2) \sigma^z$, the bath Hamiltonian is $$H_B = \sum_k \omega_k a_k^\dagger a_k,$$ and the interaction is $H_{SB} = \sigma^z X$, where $X$ is the collective bath coordinate
$$ X = \sum_k \chi_k(a_k+a_k^\dagger).$$
This is known as an independent boson model and is exactly solvable (see Emilio Pisanty's elegant general solution for the time-independent problem here), by close analogy with the solvable classical noise model discussed here.
Before presenting the solution, let us motivate the idea of replacing the quantum bath with a noisy classical field, as follows. Moving to an interaction picture generated by $H_S + H_B$, the Hamiltonian transforms to
$$ H_{SB}(t) = \sigma^z X(t),$$
where (with $\hbar = 1$)
$$X(t) = \sum_k \chi_k ( \ee^{-\ii\omega_k t} a_k + \ee^{\ii\omega_k t} a_k^\dagger ).$$
Now, let's assume that the reservoir is initially in a thermal state $\rho_B = \ee^{-\beta H_B}/\mathcal{Z}$, with $\beta = 1/k_B T$ the inverse temperature and $\mathcal{Z} = \Tr[\ee^{-\beta H_B}]$ the partition function. Then, $X(t)$ has zero mean, $\langle X(t) \rangle = \Tr[X(t) \rho_B] = 0$. However, its fluctuations are non-vanishing, as quantified by the autocorrelation function
$$ \langle X(t) X(t') \rangle = \int_{-\infty}^\infty \dd \omega\; \ee^{-\ii\omega (t-t')} S(\omega).$$
Here the quantity $S(\omega)$ is defined by
$$
S(\omega) = \left \lbrace
\begin{array}{ll}
J(\omega)[1 + n(\omega)] & (\omega > 0)\\
J(\lvert\omega\rvert) n(\lvert\omega\rvert) & (\omega < 0),
\end{array}\right.
$$
where $n(\omega) =(\ee^{\beta\omega} - 1)^{-1}$ is the Bose-Einstein distribution and the spectral density is
$$ J(\omega) = \sum_k \lvert\chi_k\rvert^2 \delta(\omega- \omega_k).$$
(Note that really $S(\omega)$, not $J(\omega)$, plays the role analogous to the noise power spectral density, but unfortunately this terminology has been cemented by several decades of repetition and we're pretty much stuck with it.) In addition, the statistics of $X(t)$ are Gaussian, in the sense that all higher-order cumulants with respect to the initial state $\rho_B$ vanish. Thus, if we imagine replacing $X(t)$ in the above Hamiltonian with a classical zero-mean noisy field having Gaussian statistics and power spectral density $S(\omega)$, we might hope to reproduce the dynamics of the full quantum model. However, this idea neglects the fact that the quantum bath is a dynamical system which is perturbed by its interaction with the qubit, and so we cannot expect the noise statistics to remain independent of the qubit's state, nor even to remain Gaussian, as time progresses. Interestingly, for the independent boson model the statistics do in fact remain Gaussian, because the qubit simply acts to displace the equilibrium position of each bosonic mode, which in turn leads the qubit to decohere without affecting its mean energy.
So, on to the solution for the qubit's reduced density matrix $\rho_S(t) = \Tr_B[U(t) \rho(0) U^\dagger(t)]$, where $U(t)$ is the time evolution operator (in the interaction picture), assuming that the initial state is the (tensor) product $\rho(0) = \rho_S(0)\rho_B$. In order to compute the time evolution operator, the key observation is that $[X(t),X(t')]$ is a $c$-number which commutes with everything. Hence, using the Magnus expansion, one finds that
\begin{align} U(t) & = T\exp \left (-\ii \int_0^t\dd s\; H_{SB}(s) \right) \\ &= \exp\left(-\ii \int_0^t\dd s\; H_{SB}(s) -\frac{1}{2}\int_0^t\dd s\int_0^s\dd s'\;[H_{SB}(s),H_{SB}(s')] \right),
\end{align}
i.e. on the second line we have reexpressed the intractable time-ordered exponential in terms of a normal exponential which can be readily evaluated. The second term in the exponent involving a commutator yields merely a negligible global phase. Dropping this phase factor, we obtain
\begin{align}
U(t) &= \exp\left\{\frac{1}{2}\sigma^z \sum_k \left[ \alpha_k(t) a_k^\dagger - \alpha^*_k(t) a_k \right]\right\},
\end{align}
where
$$ \alpha_k(t) = \frac{2\chi_k(1 - \ee^{\ii\omega_k t})}{\omega_k}.$$
Thus, $U(t)$ describes a time-dependent displacement of each mode by an amount $\pm\alpha_k(t)/2$, conditioned on the state of the qubit. Explicitly, we can write
$$ U(t) = \lvert 1 \rangle \langle 1 \rvert \prod_k D(\alpha_k/2) + \lvert 0 \rangle \langle 0 \rvert \prod_k D(-\alpha_k/2),$$
where $D(\alpha_k/2)$ denotes a standard displacement operator acting on the $k^{{\rm th}}$ mode.
Since $[H,\sigma^z] = 0$, the populations in the eigenbasis of $\sigma^z$ are constant, while the coherences decay as $\rho_{01}(t) = \langle 0 \rvert \rho_S(t) \rvert 1\rangle = \ee^{-\Gamma(t)}\rho_{01}(0)$, where we defined the decoherence function via
$$ \ee^{-\Gamma(t)} = \left\langle \prod_k D(\alpha_k)\right\rangle,$$
where the angle brackets denote an average with respect to the initial bath state $\rho_B$. Now, one can recognise the expectation value of the displacement operator as the Wigner characteristic function of the Gaussian bath state $\rho_B$. This characteristic function can be simply looked up, or otherwise derived by noting that $\Gamma(t)$ is essentially the cumulant generating function of the variable $\sum_k(\alpha_k a_k^\dagger - \alpha_k^* a_k)$, and for a Gaussian state only its second cumulant contributes, therefore
\begin{align} \Gamma(t) &= \sum_k \frac{4 \lvert\chi_k\rvert^2}{\omega_k^2} [1-\cos(\omega_k t)]\coth \left(\beta\omega_k/2\right)\\
& = \int_{-\infty}^\infty \dd\omega\; \frac{4[1-\cos(\omega t)]}{\omega^2} S(\omega). \end{align}
Alternatively, we can write these results in terms of a master equation in the Schroedinger picture:
$$ \dot{\rho} = -\frac{\ii}{2}\omega_0 [\sigma^z,\rho] + \frac{1}{2}\gamma(t)\left( \sigma^z \rho \sigma^z - \rho \right). $$
Here, dots denote time derivatives and the dephasing rate is $\gamma(t) = \dot{\Gamma}(t)$, or explicitly
$$ \gamma(t) = \int\dd\omega\; \frac{\sin(\omega t)}{\omega} S(\omega).$$
Comparison of these results with the answer here shows that the evolution is equivalent to that produced by a classical noisy field with stationary Gaussian statistics and power spectral density $S(\omega)$. However, actually generating such a classical noisy signal seems challenging, since $S(\omega)$ is not an even function of $\omega$ as it would be for a real-valued classical signal. This suggests that such a classical noise approximation makes sense only in the high-temperature limit, where $n(\omega) \gg 1$, since $S(\omega)$ becomes approximately even for frequencies $\lvert \beta \omega\rvert \ll 1$.
