Since the top-dimensional exterior power is one-dimensional, we may talk about the quotient $\alpha/\beta \in C^1(\mathscr{M})$ of two volume forms.
The divergence only depends on the volume form of $g$, and we have the identity
$$
\operatorname{div}_{g'}(X) = \frac{\omega_{g\hphantom{}}}{\omega_{g'}} \,\operatorname{div}_{g}\left(\frac{\omega_{g}}{\omega_{g'}} \cdot X\right),
$$
where $\omega_g$ and $\omega_{g}'$ are the volume forms of $g$ and $g'$, respectively. For what follows, define $\rho_{g, g'} \in C^1(\mathscr{M})$ by $\rho_{g, g'} = \omega_g/\omega_{g'} - 1$ and define $\lambda_{g,g'} \in C^1(\mathrm{T}\mathscr{M})$ by $\lambda_{g, g'}(v) = |v|_{g}/|v|_{g'}$.
Then for any $C^1$ vector field $X$, we have $\int_{\Omega} \operatorname{div}_{g'} (X)\,\omega_{g'} - \int_{\Omega} \operatorname{div}_{g}(X)\, \omega_g = \int_{\Omega} \operatorname{div}_{g} \left(\rho_{g, g'} X \right)\,\omega_g$; furthermore we have
\begin{equation*}
\int_{\Omega} \operatorname{div}_{g} (X)\,\omega_{g} = \sup_{\Omega} (\lambda_{g, g'} \circ X) \underbrace{\int_{\Omega} \operatorname{div}_{g} \left(\frac{X}{\sup_{\Omega} (\lambda_{g,g'} \circ X)} \right)\,\omega_{g}}_{\leq \,\operatorname{Per}_{g}(\Omega) \text{ if } \lVert X \rVert_{g'} \leq 1}
\end{equation*}
and because $\lambda_{g, g'}$ is invariant under scaling,
\begin{align*}
\int_{\Omega} \operatorname{div}_{g} \left(\rho_{g, g'} X \right)\,\omega_g &= \sup_{\Omega} |\rho_{g, g'}| \int_{\Omega} \operatorname{div}\left(\frac{\rho_{g, g'}X}{\sup_{\Omega} |\rho_{g,g'}| } \right)\,\omega_{g} \\[1ex]
&= \sup_{\Omega} (\lambda_{g, g'} \circ X) \sup_{\Omega} |\rho_{g, g'}|\! \underbrace{\int_{\Omega} \operatorname{div}_{g} \left(\frac{\rho_{g, g'} X}{\sup_{\Omega} (\lambda_{g, g'} \circ X) \cdot \sup_{\Omega} |\rho_{g,g'}|} \right)\,\omega_g}_{\leq \, \operatorname{Per}_{g}(\Omega) \text{ if } \lVert X \rVert_{g'} \leq 1}.
\end{align*}
Let $\mathrm{U}\mathscr{M}(g') \subseteq \mathrm{T}\mathscr{M}$ be the unit tangent bundle with respect to $g'$, and define $\Lambda_{g, g'} = \sup_{v \in \mathrm{U}\mathscr{M}(g')} |v|_{g}$. Since $\mathrm{U}\mathscr{M}(g')$ is compact, $\Lambda_{g,g'}$ is finite. It follows that
$$
\sup_{\Omega} \lambda_{g, g'} X = \sup_{\Omega} \left|\frac{\,X_{\hphantom{g'}}\!}{|X|_{g'}\!}\right|_{\raise{1.5pt}{g}} \leq \Lambda_{g, g'}.
$$
It remains to show that $\lim_{g' \to g} \Lambda_{g, g'} = 1$ and $\lim_{g' \to g}\sup_{\Omega} |\rho_{g, g'}| = 0$. To do this, we use the following characterization of $C^0$ convergence:
Proposition. Let $(s_k)_{k \in \mathbb{N}}$ be a sequence of sections of $\mathrm{T}^*\mathscr{M} \otimes \mathrm{T}^*\mathscr{M}$. Then $(s_k)_{k \in \mathbb{N}}$ uniformly converges to a section $s$ if and only if for every pair of vector fields $X, Y \in \Gamma(\mathrm{T}\mathscr{M})$, we have $s_k(X, Y) \to s(X, Y)$ uniformly.
To use this, first note that there exists an open cover $\{U_\alpha\}_{\alpha \in A}$ of $\mathscr{M}$ such that for each $\alpha \in A$, there exists an orthonormal frame $\xi^\alpha \in \mathrm{F}U_\alpha (g)$.
To show that $\lim_{g' \to g} \Lambda_{g, g'} = 1$, take a convergent sequence $g_k \to g$, and let $\epsilon > 0$. Then there exists some index $K$ such that for all $k \geq K$ and $\alpha \in A$, we have $\lVert \delta_{ij} - g_k(\xi^\alpha_i, \xi^\alpha_j) \rVert_{C^0} < \epsilon$.
Let $x \in \mathscr{M}$; this point must belong to some $U_{\alpha}$. Let $v \in \mathrm{U}_x\mathscr{M}(g_k)$, and write $v^i = g(v, \xi^\alpha_i|_x)$. Then
$$
|v|_{g_k}^2 = \left| \sum_{i=1}^{n} v^i \xi^\alpha_i|_x \right|_{g_k} = \sum_{i, j} g_k(\xi^\alpha_i, \xi^\alpha_j) v^iv^j = 1.
$$
In particular, if $\Sigma_{k, x}$ are the eigenvalues of $g_k|_x$ in this frame, then $|v|_{g}^2$ is contained in the interval $[(\max \Sigma_{k,x})^{-1}, (\min \Sigma_{k, x})^{-1}]$. However, by the Gershgorin circle theorem,
$$\Sigma_{k, x} \subseteq \left[1 - \max_{i} \sum_{j = 1}^{n} |g_k(\xi^\alpha_i, \xi^\alpha_j)|, 1 + \max_{i} \sum_{j = 1}^{n} |g_k(\xi^\alpha_i, \xi^\alpha_j)|\right] \subseteq \left[1 - n\epsilon, 1 + n\epsilon \right].$$
It follows that $|1 - \sup_{v \in \mathrm{U}\mathscr{M}(g_k)} |v|_{g}|$ is close to $0$. Therefore, $\lim_{g' \to g} \Lambda_{g, g'} = 1$.
To show that $\lim_{g' \to g} \sup_\Omega |\rho_{g, g'}| = 0$, take a convergent sequence $g_k \to g$, and let $\epsilon > 0$. As done before, let $x \in \mathscr{M}$; this point must belong to some $U_{\alpha}$. Then
$$
\rho_{g, g_k}(x)\hspace{-1pt} = \hspace{-1pt} 1 - \hspace{-1pt} \frac{\omega_{g}(\xi^{\alpha}_1 \wedge \cdots \wedge \xi^{\alpha}_n)}{\omega_{g_k}(\xi^{\alpha}_1 \wedge \cdots \wedge \xi^{\alpha}_n)} = \hspace{-1pt} 1 - \frac{1}{\omega_{g_k}(\xi^{\alpha}_1 \wedge \cdots \wedge \xi^{\alpha}_n)} = \hspace{-1pt} 1 - \frac{1}{\sqrt{\det \big[g_k(\xi^\alpha_i, \xi^{\alpha_j}) \big]_{ij}}}.
$$
Applying the Gershgorin circle theorem again, we obtain $\Sigma_{k,x} \subseteq [1-\epsilon, 1 + \epsilon]$. Hence, the product of the eigenvalues is also close to $1$. It follows that $\lim_{g' \to g} \sup_\Omega |\rho_{g, g'}| = 0$.
The conclusion now follows. First take a sequence of vector fields $X_i \in C^1(\mathscr{M}, \mathrm{T}\mathscr{M})$ which realizes the perimeter of $\Omega$ with respect to $g'$. Then
$$
\operatorname{Per}_{g'}(\Omega) \leq \Lambda_{g, g'} \operatorname{Per}_{g}(\Omega) + \Lambda_{g, g'} \sup_{\Omega} |\rho_{g, g'}| \operatorname{Per}_{g}(\Omega).
$$
Therefore, if $g_k \to g$ uniformly, then $\limsup_{k \to \infty} \operatorname{Per}_{g_k} (\Omega) \leq \operatorname{Per}_{g}(\Omega)$. Thus the perimeter is upper semi-continuous, and hence continuous.
