Yes "Riemann", your hypothesis is true.  

Landau showed the count of all semiprimes grows as follows:
$$
|\{pq \leq x\}| \sim \frac{x}{\log x}(\log \log x).
$$

If $\gcd(a,n) = 1$ then a special case of the answer by Lucia [here][1] says 
$$
|\{pq \leq x : pq \equiv a \bmod n\}| \sim 
\frac{1}{\varphi(n)}\frac{x}{\log x}(\log \log x)
$$
because Lucia's result implies that, for each $b, c \in (\mathbf Z/n\mathbf Z)^\times$, the number of semiprimes $pq \leq x$ where $p \equiv b  \bmod n$ and $q \equiv c \bmod n$ is asymptotic to 
$(1/\varphi(n)^2)(x/\log x)(\log\log x)$. You have to be careful about avoiding duplicate counting of the products $pq$ depending on whether or not $b \equiv c \bmod n$. Therefore 
\begin{align*}
|\{pq \leq x : pq \equiv a \bmod n\}| & = 
\sum_{\substack{(b,c) \bmod n \\ bc \equiv a \bmod n}} |\{pq \leq x : p \equiv b \bmod n, q \equiv c \bmod n\}| \\
& = 
\sum_{\substack{b \bmod n}} |\{pq \leq x : p \equiv b \bmod n, q \equiv b^{-1}a \bmod n\}| \\
& \sim 
\sum_{\substack{b \bmod n}} \frac{1}{\varphi(n)^2}\frac{x}{\log x}(\log \log x) \\
& = \frac{1}{\varphi(n)}\frac{x}{\log x}(\log \log x). 
\end{align*}


There is a more general result. For fixed $k \geq 1$, Landau showed 
$$
|\{p_1\cdots p_k \leq x\}| \sim \frac{x}{\log x}\frac{(\log \log x)^{k-1}}{(k-1)!}
$$
(that is counting $k$-almost primes $p_1\cdots p_k$, not the $k$-tuples $(p_1,\ldots,p_k)$, so order and multiplicity matter when passing between these two methods of counting), and one then expects when $\gcd(a,n) = 1$ that 
$$
|\{p_1\cdots p_k \leq x : p_1\cdots p_k \equiv a \bmod n\}| \sim 
\frac{1}{\varphi(n)}\frac{x}{\log x}\frac{(\log \log x)^{k-1}}{(k-1)!}.
$$
Such an estimate follows from Lucia's result by estimating 
$$
|\{p_1\cdots p_k \leq x : p_1 \equiv b_1 \bmod n, \ldots, p_k \equiv b_k \bmod n\}|
$$
and summing those estimates over the $\varphi(n)^{k-1}$ different $k$-tuples 
$(b_1,\ldots,b_k)$ from $(\mathbf Z/n\mathbf Z)^\times$ where
$b_1\cdots b_k \equiv a \bmod n$ (let $b_1, \ldots, b_{k-1}$ be arbitrary units mod $n$ and then $b_k \bmod n$ is determined by the congruence $b_1\cdots b_k \equiv a \bmod n$).

[1]:https://mathoverflow.net/questions/156982/chebotarev-density-theorem-for-k-almost-primes