Let $P_1$, $\ldots$, $P_r$ be disjoint subsets of the primes such that the asymptotic formula $$ \sum_{\substack{{p\le x}\\ {p\in P_j}}} 1 \sim \alpha_j \frac{x}{\log x} $$ holds with some $\alpha_j >0$. Let $N_k=N(k;a_1,\ldots,a_r)$ denote the set of integers that are products of $k$ primes with exactly $a_j$ of these primes chosen from the set $P_j$. Thus $a_1+\ldots+a_r=k$ and let's assume that all $a_j \ge 1$.
I claim that for fixed $k$ and as $x\to \infty$ $$ \sum_{\substack {{n\le x}\\ {n\in N_k}}} 1 \sim k\prod_{j=1}^{r} \frac{\alpha_j^{a_j}}{(a_j)!} \frac{x}{(\log x)} (\log \log x)^{k-1} $$ The argument that follows is standard. The case $k=1$ follows from our assumption, and now suppose that $k\ge 2$.

Put $z=x^{1/\log \log x}$ and note that by partial summation $$ \sum_{p\in P_j, p\le z} \frac{1}{p} \sim \alpha_j \log \log z \sim \alpha_j \log \log x $$ while $$ \sum_{p \in P_j, z< p \le x} \frac{1}{p} \sim \alpha_j \log \log \log x. $$ Now suppose $n=p_1\cdots p_k$ is an element of $N(k;a_1,\ldots,a_r)$. We distinguish two cases: when the largest prime factor of $n$ is bigger than $z$ but all the other prime factors are smaller than $z$, and when the two largest prime factors of $n$ are both larger than $z$. (Of course there is also the case when all prime factors of $n$ are smaller than $z$, but this is clearly negligible.) The first case will be the main term and the second case will be a slightly smaller error.

Let's look at the first case. Suppose the largest prime factor (say $p$) of $n$ lies in the set $P_j$, so that the product of the smaller primes, say $m$, lies in $N(k-1;a_1,\ldots,a_{j}-1,\ldots,a_r)$. These terms give, upon summing over the last prime $p$, (note that as $m\le z^{k-1}$ we have $\log x \sim \log (x/m)$)
$$ \sim \sum_{\substack{{m\in N(k-1;\ldots)} \\ {p|m\implies p\le z}} } \frac{\alpha_j x}{m \log (x/m)} \sim \frac{\alpha_j x}{\log x} \prod_{t=1, t\neq j}^{r} \frac{1}{a_t!} \Big(\sum_{p\le z, p\in P_t} \frac 1p\Big)^{a_t} \frac{1}{(a_j-1)!} \Big(\sum_{p\le z, p\in P_j}\frac 1p\Big)^{a_j-1}. $$ This is $$ \sim a_j \prod_{t=1}^{r} \frac{\alpha_t^{a_t}}{a_t!} \frac{x}{\log x} (\log \log x)^{k-1} $$ and summing it over $j$ from $1$ to $r$ gives the claimed asymptotic.

Now for the second case. The sum the largest prime $p$ gives a quantity bounded by (let $m$ be the product of the remaining primes and note that $m\le x^{(k-1)/k}$) $$ \frac{x/m}{\log (x/m)} \ll \frac{x}{m \log x}. $$ Now sum this over the possible values of $m$, keeping in mind that $m$ has $k-1$ prime factors, and the largest prime factor of $m$ is assumed to lie in $[z,x]$.
Thus these terms are bounded by $$ \ll \frac{x}{\log x} \Big(\sum_{p\le x} \frac{1}{p} \Big)^{k-2} \Big(\sum_{z< p\le x} \frac{1}{p}\Big) \ll \frac{x}{(\log x)} (\log \log x)^{k-2} (\log \log \log x). $$

This should now definitively answer your Chebotarev question (assuming I got my combinatorics right) at least when $k$ is fixed. When $k$ grows, one has to do more work, but the Selberg-Delange method should work in this case with some effort.

I'll prove a more general result from which the Chebotarev question will follow. Let $P_1$, $\ldots$, $P_r$ be disjoint subsets of the primes such that the asymptotic formula $$ \sum_{\substack{{p\le x}\\ {p\in P_j}}} 1 \sim \alpha_j \frac{x}{\log x} $$ holds with some $\alpha_j >0$. Let $N_k=N(k;a_1,\ldots,a_r)$ denote the set of integers that are products of $k$ primes with exactly $a_j$ of these primes chosen from the set $P_j$. Thus $a_1+\ldots+a_r=k$ and let's assume that all $a_j \ge 1$.
I claim that for fixed $k$ and as $x\to \infty$ $$ \sum_{\substack {{n\le x}\\ {n\in N_k}}} 1 \sim k\prod_{j=1}^{r} \frac{\alpha_j^{a_j}}{(a_j)!} \frac{x}{(\log x)} (\log \log x)^{k-1} $$ The argument that follows is standard. The case $k=1$ follows from our assumption, and now suppose that $k\ge 2$.

Put $z=x^{1/\log \log x}$ and note that by partial summation $$ \sum_{p\in P_j, p\le z} \frac{1}{p} \sim \alpha_j \log \log z \sim \alpha_j \log \log x $$ while $$ \sum_{p \in P_j, z< p \le x} \frac{1}{p} \sim \alpha_j \log \log \log x. $$ Now suppose $n=p_1\cdots p_k$ is an element of $N(k;a_1,\ldots,a_r)$. We distinguish two cases: when the largest prime factor of $n$ is bigger than $z$ but all the other prime factors are smaller than $z$, and when the two largest prime factors of $n$ are both larger than $z$. (Of course there is also the case when all prime factors of $n$ are smaller than $z$, but this is clearly negligible.) The first case will be the main term and the second case will be a slightly smaller error.

Let's look at the first case. Suppose the largest prime factor (say $p$) of $n$ lies in the set $P_j$, so that the product of the smaller primes, say $m$, lies in $N(k-1;a_1,\ldots,a_{j}-1,\ldots,a_r)$. These terms give, upon summing over the last prime $p$, (note that as $m\le z^{k-1}$ we have $\log x \sim \log (x/m)$)
$$ \sim \sum_{\substack{{m\in N(k-1;\ldots)} \\ {p|m\implies p\le z}} } \frac{\alpha_j x}{m \log (x/m)} \sim \frac{\alpha_j x}{\log x} \prod_{t=1, t\neq j}^{r} \frac{1}{a_t!} \Big(\sum_{p\le z, p\in P_t} \frac 1p\Big)^{a_t} \frac{1}{(a_j-1)!} \Big(\sum_{p\le z, p\in P_j}\frac 1p\Big)^{a_j-1}. $$ This is $$ \sim a_j \prod_{t=1}^{r} \frac{\alpha_t^{a_t}}{a_t!} \frac{x}{\log x} (\log \log x)^{k-1} $$ and summing it over $j$ from $1$ to $r$ gives the claimed asymptotic.

Now for the second case. The sum over the largest prime $p$ gives a quantity bounded by (let $m$ be the product of the remaining primes and note that $m\le x^{(k-1)/k}$) $$ \frac{x/m}{\log (x/m)} \ll \frac{x}{m \log x}. $$ Now sum this over the possible values of $m$, keeping in mind that $m$ has $k-1$ prime factors, and the largest prime factor of $m$ is assumed to lie in $[z,x]$.
Thus these terms are bounded by $$ \ll \frac{x}{\log x} \Big(\sum_{p\le x} \frac{1}{p} \Big)^{k-2} \Big(\sum_{z< p\le x} \frac{1}{p}\Big) \ll \frac{x}{(\log x)} (\log \log x)^{k-2} (\log \log \log x). $$

Application to the Chebotarev question. Suppose $C_1$, $\ldots$, $C_r$ are distinct conjugacy classes in $G$. Suppose each element of the set $D$ (which is a $k$-tuple) consists of $a_1$ entries from $C_1$, $a_2$ entries from $C_2$, $\ldots$, $a_r$ entries from $C_r$. A little combinatorics shows that the size of the set $D$ (assumed to be conjugation and permutation invariant) is $$ |D| = k!\prod_{j=1}^{r} \frac{|C_j|^{a_j}}{a_j!}. $$ In our work above take $P_j$ to be the primes $p$ for which Frobenius lies in the class $C_j$. Then by Chebotarev our assumed asymptotic for the primes in $P_j$ holds with $\alpha_j=|C_j|/|G|$. It follows that the number of integers in the original question is $$ \sim \frac{|D|}{|G|^k} \frac{x}{\log x} \frac{(\log \log x)^{k-1}}{(k-1)!}. $$ This answers the Chebotarev question (assuming I got my combinatorics right) at least when $k$ is fixed. When $k$ grows, one has to do more work, but the Selberg-Delange method should work in this case with some effort; note here that one needs to use the structure of sets of Chebotarev primes, rather than just an arbitrary subset of the primes as above.

Let $P$ be a subset of the primes such that the asymptotic formula $$ \sum_{\substack{{p\le x}\\ {p\in P}}} 1 \sim \alpha \frac{x}{\log x} $$ holds with some $\alpha >0$. Let $P_k$ denote the set of integers that are products of $k$ primes from $P$ (let's say distinct for simplicity). I claim that for fixed $k$ and as $x\to \infty$ $$ \sum_{\substack {{n\le x}\\ {n\in P_k}}} 1 \sim \alpha^k \frac{x}{(\log x)} \frac{(\log \log x)^{k-1}}{(k-1)!}. $$ The argument that follows is standard. The case $k=1$ is our assumption, and now suppose that $k\ge 2$.

Put $z=x^{1/\log \log x}$ and note that by partial summation $$ \sum_{p\in P, p\le z} \frac{1}{p} \sim \alpha \log \log z \sim \alpha \log \log x $$ while $$ \sum_{p \in P, z< p \le x} \frac{1}{p} \sim \alpha \log \log \log x. $$ Now suppose $n=p_1\cdots p_k$ is an element of $P_k$ below $x$ with the primes $p_j$ in ascending order. We distinguish two cases: when $p_{k-1}<z$ and when $p_{k-1}>z$. The first case will be the main term and the second case will be a slightly smaller error. Let's look at the first case. These terms give, upon summing over the last prime $p_k$, (note that as $p_1 <\ldots <p_{k-1}<z$ we have $\log x \sim \log (x/p_1\cdots p_{k-1})$)
$$ \sim \sum_{p_1<\ldots<p_{k-1} < z} \frac{\alpha x}{p_1\cdots p_{k-1} \log (x/(p_1\cdots p_{k-1}))} \sim \frac{\alpha x}{\log x} \frac{1}{(k-1)!} \Big(\sum_{p\in P, p\le z} \frac{1}{p}\Big)^{k-1}, $$ which gives the claimed asymptotic.

Now for the second case. The sum over $p_{k}$ gives a quantity bounded by $$ \alpha \frac{x/(p_1\cdots p_{k-1})}{\log (x/p_1 \cdots p_{k-1})} \ll \frac{x}{(p_1\cdots p_{k-1}) \log x} $$ since we may assume that $p_1\cdots p_{k-1} \le x^{(k-1)/k}$ else the sum over $p_k$ would be empty.
Now sum this over the possible values of $p_j$, keeping in mind that $p_{k-1}$ is assumed to lie in $[z,x]$.
Thus these terms are bounded by $$ \ll \frac{x}{\log x} \Big(\sum_{p\le x, p\in P} \frac{1}{p} \Big)^{k-2} \Big(\sum_{z< p\le x, p\in P} \frac{1}{p}\Big) \ll \frac{x}{(\log x)} (\log \log x)^{k-2} (\log \log \log x). $$

Thus in particular your Chebotarev question holds with $c$ there being $(|D|/|G|)^{k}$. When $k$ grows, one has to do more work, but the Selberg-Delange method should work in this case with some effort.

Let $P_1$, $\ldots$, $P_r$ be disjoint subsets of the primes such that the asymptotic formula $$ \sum_{\substack{{p\le x}\\ {p\in P_j}}} 1 \sim \alpha_j \frac{x}{\log x} $$ holds with some $\alpha_j >0$. Let $N_k=N(k;a_1,\ldots,a_r)$ denote the set of integers that are products of $k$ primes with exactly $a_j$ of these primes chosen from the set $P_j$. Thus $a_1+\ldots+a_r=k$ and let's assume that all $a_j \ge 1$.
I claim that for fixed $k$ and as $x\to \infty$ $$ \sum_{\substack {{n\le x}\\ {n\in N_k}}} 1 \sim k\prod_{j=1}^{r} \frac{\alpha_j^{a_j}}{(a_j)!} \frac{x}{(\log x)} (\log \log x)^{k-1} $$ The argument that follows is standard. The case $k=1$ follows from our assumption, and now suppose that $k\ge 2$.

Put $z=x^{1/\log \log x}$ and note that by partial summation $$ \sum_{p\in P_j, p\le z} \frac{1}{p} \sim \alpha_j \log \log z \sim \alpha_j \log \log x $$ while $$ \sum_{p \in P_j, z< p \le x} \frac{1}{p} \sim \alpha_j \log \log \log x. $$ Now suppose $n=p_1\cdots p_k$ is an element of $N(k;a_1,\ldots,a_r)$. We distinguish two cases: when the largest prime factor of $n$ is bigger than $z$ but all the other prime factors are smaller than $z$, and when the two largest prime factors of $n$ are both larger than $z$. (Of course there is also the case when all prime factors of $n$ are smaller than $z$, but this is clearly negligible.) The first case will be the main term and the second case will be a slightly smaller error.

Let's look at the first case. Suppose the largest prime factor (say $p$) of $n$ lies in the set $P_j$, so that the product of the smaller primes, say $m$, lies in $N(k-1;a_1,\ldots,a_{j}-1,\ldots,a_r)$. These terms give, upon summing over the last prime $p$, (note that as $m\le z^{k-1}$ we have $\log x \sim \log (x/m)$)
$$ \sim \sum_{\substack{{m\in N(k-1;\ldots)} \\ {p|m\implies p\le z}} } \frac{\alpha_j x}{m \log (x/m)} \sim \frac{\alpha_j x}{\log x} \prod_{t=1, t\neq j}^{r} \frac{1}{a_t!} \Big(\sum_{p\le z, p\in P_t} \frac 1p\Big)^{a_t} \frac{1}{(a_j-1)!} \Big(\sum_{p\le z, p\in P_j}\frac 1p\Big)^{a_j-1}. $$ This is $$ \sim a_j \prod_{t=1}^{r} \frac{\alpha_t^{a_t}}{a_t!} \frac{x}{\log x} (\log \log x)^{k-1} $$ and summing it over $j$ from $1$ to $r$ gives the claimed asymptotic.

Now for the second case. The sum the largest prime $p$ gives a quantity bounded by (let $m$ be the product of the remaining primes and note that $m\le x^{(k-1)/k}$) $$ \frac{x/m}{\log (x/m)} \ll \frac{x}{m \log x}. $$ Now sum this over the possible values of $m$, keeping in mind that $m$ has $k-1$ prime factors, and the largest prime factor of $m$ is assumed to lie in $[z,x]$.
Thus these terms are bounded by $$ \ll \frac{x}{\log x} \Big(\sum_{p\le x} \frac{1}{p} \Big)^{k-2} \Big(\sum_{z< p\le x} \frac{1}{p}\Big) \ll \frac{x}{(\log x)} (\log \log x)^{k-2} (\log \log \log x). $$

This should now definitively answer your Chebotarev question (assuming I got my combinatorics right) at least when $k$ is fixed. When $k$ grows, one has to do more work, but the Selberg-Delange method should work in this case with some effort.