Suppose that $Ju_i$ does not converge to $Ju$ in $L^1$. Then for a subsequence you will have $\Vert Ju_i-Ju\Vert_1\geq\epsilon$. If you take a subequence you can assume that $\nabla u_i\to\nabla u$ a.e. and hence $Ju_i\to Ju$ a.e. which would imply convergence $Ju_i\to Ju$ on that sequence and that would contradict the inequality $\Vert Ju_i-Ju\Vert_1\geq\epsilon$.

To make thigs more straightforward, you can also use a finite covering of $\mathcal{M}$ so you can assume that $u,u_i\in W^{1,n}(B^n(0,1),\mathcal{N})$.

In fact your claim is true not only for Lipschitz maps but for any $W^{1,n}$.

You actually do not need to assume that the mappings are Lipschitz as it is true for general $W^{1,n}$ mappings

Theorem. If $\mathcal{M}$ and $\mathcal{N}$ are smooth compact and oriented manifolds, $\mathcal{N}\subset\mathbb{R}^D$, and $u,u_k\in W^{1,n}(\mathcal{M},\mathcal{N})$, $u_k\to u$ in $W^{1,n}$, then the Jacobians converge in $L^1$, $Ju_k\to Ju$.

Proof. Suppose that $Ju_k$ does not converge to $Ju$ in $L^1$. Then for a subsequence (still denoted by $u_k$) we will have $\Vert Ju_k-Ju\Vert_1\geq\epsilon$. If we take a further subequence, we can also assume that $u_k\to u$ and $Du_k\to Du$ a.e.

Since $\mathcal{M}$ is compact, we can use a finite atlas which allows us to assume that $\mathcal{M}=B^n(0,1)$. Since the mappings need not be continuous, localization of the mappings in an atlas on $\mathcal{N}$ is not possible.

It is assumed that $\mathcal{N}$ is a submanifold of $\mathbb{R}^D$. Let $\omega$ be the volume form on $\mathcal{N}$. By extension, we can always assume that $\omega$ is a compactly supported smooth form on $\mathbb{R}^D$ so $$ \omega=\sum_{|I|=n}\omega_I dy^I, \quad dy^I=dy^{i_1}\wedge\ldots\wedge dy^{i_n}, \quad 1\leq i_1<\ldots<i_n\leq D. $$

If $u\in W^{1,n}(B^n(0,1),\mathcal{N})$, then we can interpret the Jacobian as the $n$-form: $$ u^*\omega(x)=\sum_{|I|}(\omega_I\circ u)du^{i_1}\wedge\ldots\wedge du^{i_n}=Ju(x)dx^1\wedge\ldots\wedge dx^n $$ Note that $\omega_I\circ u_k\to \omega_I\circ u$ a.e. and the functions are uniformly bounded because $\omega$ is bounded as a function on $\mathbb{R}^D$. Also it easily follows from the triangle inequality and Holder's inequality that $$ du_k^{i_1}\wedge\ldots\wedge du_k^{i_n}\to du^{i_1}\wedge\ldots\wedge du^{i_n} \quad \text{in $L^1$.} $$ Therefore, it easily follows (triangle inequality plus dominated convergence theorem) that $u_k^*\omega\to u^*\omega$ in $L^1$, but that contradicts $\Vert Ju_k-Ju\Vert_1\geq\epsilon$.

That is true and. I will write a proof later.

Suppose that $Ju_i$ does not converge to $Ju$ in $L^1$. Then for a subsequence you will have $\Vert Ju_i-Ju\Vert_1\geq\epsilon$. If you take a subequence you can assume that $\nabla u_i\to\nabla u$ a.e. and hence $Ju_i\to Ju$ a.e. which would imply convergence $Ju_i\to Ju$ on that sequence and that would contradict the inequality $\Vert Ju_i-Ju\Vert_1\geq\epsilon$.

To make thigs more straightforward, you can also use a finite covering of $\mathcal{M}$ so you can assume that $u,u_i\in W^{1,n}(B^n(0,1),\mathcal{N})$.

In fact your claim is true not only for Lipschitz maps but for any $W^{1,n}$.