Yes, in this case we can interchange the limits. That's most easily seen by computing the two iterated limits.
Let us look at the space $L^2([0,2\pi],\mu)$, where $\mu$ is the normalised Lebesgue measure, so that $\mu([0,2\pi]) = 1$. The system $\{e_k : k \in \mathbb{Z}\}$, where $e_k(t) = e^{ikt}$, is a Hilbert basis of that space.
By assumption, each $q_m$ is a trigonometric polynomial, so it can be written
$$q_m(t) = \sum_{k = -K_m}^{K_m} c_k^{(m)} e^{ikt}.$$
Then we have
$$q_m(nt) = \sum_{k = -K_m}^{K_m} c_k^{(m)} e^{i(nk)t}.$$
Since $\bigl(e_{k_{\ell}}\bigr)_{\ell \in \mathbb{N}}$ converges weakly to $0$ for every sequence $(k_\ell)$ in $\mathbb{Z}$ with $\lvert k_{\ell}\rvert \to \infty$ (that is essentially also known as the Riemann-Lebesgue lemma), it follows that
$$\int_0^{2\pi} f(t) q_m(nt)\,d\mu(t) = \sum_{k = -K_m}^{K_m} c_k^{(m)}\int_0^{2\pi} f(t) e^{i(nk)t}\,d\mu(t) \xrightarrow{n\to\infty} c_0^{(m)} \int_0^{2\pi} f(t)\,d\mu(t)$$
for every $f\in L^2([0,2\pi],\mu)$.
Now we have the assumption that $(q_m)$ converges uniformly to $g$ as $m \to \infty$ - that is a stronger assumption than we need, it suffices that $q_m \to g$ weakly in $L^2([0,2\pi],\mu)$ - and thus
$$c_0^{(m)} = \int_0^{2\pi} q_m(t)\,d\mu(t) \xrightarrow{m\to\infty} \int_0^{2\pi} g(t)\,d\mu(t).$$
Hence we have
$$\lim_{m\to \infty} \Biggl( \lim_{n\to \infty} \int_0^{2\pi} f(t)q_m(nt)\,d\mu(t)\Biggr) = \lim_{m\to \infty} c_0^{(m)} \int_0^{2\pi} f(t)\,d\mu(t) = \int_0^{2\pi} g(t)\,d\mu(t)\cdot \int_0^{2\pi} f(t)\,d\mu(t)$$
for every $f \in L^2([0,2\pi],\mu)$.
On the other hand, for every fixed $n$, we have by assumption uniform convergence of $q_m(nt)$ to $g(nt)$ - again, that is a stronger assumption than we need, weak convergence $q_m \rightharpoonup g$ in $L^2([0,2\pi],\mu)$ suffices - and hence
$$\lim_{m\to \infty} \int_0^{2\pi} f(t)q_m(nt)\,d\mu(t) = \int_0^{2\pi} f(t) g(nt)\,d\mu(t)$$
for every $f \in L^2([0,2\pi],\mu)$.
Now we can express $g$ through its Fourier series,
$$g = \sum_{k \in \mathbb{Z}} c_k e_k.$$
Then we have
$$g(nt) = \sum_{k \in \mathbb{Z}} c_k e_{nk}(t),$$
and
$$\int_0^{2\pi} f(t)g(nt)\,d\mu(t) = \sum_{k \in \mathbb{Z}} c_k \int_0^{2\pi} f(t) e_{nk}(t)\,d\mu(t).$$
Since
$$\lim_{n\to \infty} \int_0^{2\pi} f(t) e_{nk}(t)\,d\mu(t) = 0$$
for all $k \in \mathbb{Z}\setminus \{0\}$, it won't come as a surprise that
$$\lim_{n\to \infty} \int_0^{2\pi} f(t)g(nt)\,d\mu(t) = c_0\int_0^{2\pi} f(t)\,d\mu(t) = \int_0^{2\pi} g(t)\,d\mu(t) \cdot \int_0^{2\pi} f(t)\,d\mu(t).$$
But since we have (potentially) infinitely many nonzero coefficients $c_k$, we must argue a little more carefully. Since
$$\lVert g\rVert^2 = \sum_{k \in \mathbb{Z}} \lvert c_k\rvert^2 < +\infty,$$
for every given $\varepsilon > 0$ we can choose a $K_{\varepsilon}$ such that
$$\sum_{\lvert k\rvert > K_{\varepsilon}} \lvert c_k\rvert^2 < \varepsilon^2.$$
Then we write
$$g = \underbrace{\sum_{\lvert k\rvert \leqslant K_{\varepsilon}} c_k e_k}_{g_1} + \underbrace{\sum_{\lvert k\rvert > K_{\varepsilon}} c_k e_k}_{g_2},$$
and have
$$\lVert g_2(nt)\rVert = \Biggl(\sum_{\lvert k\rvert > K_{\varepsilon}} \lvert c_k\rvert^2\Biggr)^{\frac{1}{2}} = \lVert g_2\rVert < \varepsilon,$$
whence by the Cauchy-Schwarz inequality
$$\Biggl\lvert \int_0^{2\pi} f(t) g_2(nt)\,d\mu(t)\biggr\rvert \leqslant \lVert f\rVert\cdot \lVert g(nt)\rVert < \varepsilon\lVert f\rVert$$
independently of $n$. For the trigonometric polynomial $g_1$ we have
$$\lim_{n\to\infty} \int_0^{2\pi} f(t) g_1(nt)\,d\mu(t) = c_0 \int_0^{2\pi} f(t)\,d\mu(t)$$
as above, and therefore
\begin{align}
\limsup_{n\to\infty} \Biggl\lvert \int_0^{2\pi} f(t)g(nt)\,d\mu(t) &\:- c_0 \int_0^{2\pi} f(t)\,d\mu(t)\Biggr\rvert\\
&\leqslant \limsup_{n\to\infty} \Biggl\lvert \int_0^{2\pi} f(t)g_1(nt)\,d\mu(t) - c_0\int_0^{2\pi} f(t)\,d\mu(t)\Biggr\rvert\\
&\quad + \limsup_{n\to\infty} \Biggl\lvert \int_0^{2\pi} f(t)g_2(nt)\,d\mu(t)\Biggr\rvert\\
&< 0 + \varepsilon \lVert f\rVert.
\end{align}
Since this holds for all $\varepsilon > 0$, it follows that indeed
$$\lim_{n\to \infty} \Biggl( \lim_{m\to \infty} \int_0^{2\pi} f(t)q_m(nt)\,d\mu(t)\Biggr) = c_0\int_0^{2\pi} f(t)\,d\mu(t) = \int_0^{2\pi} g(t)\,d\mu(t)\cdot \int_0^{2\pi} f(t)\,d\mu(t),$$
as desired. We note that $f$ can be an arbitrary $L^2([0,2\pi],\mu)$ function, and that we don't need uniform convergence $q_m \to g$, weak convergence in $L^2([0,2\pi],\mu)$ suffices.
