$\newcommand{\CO}{\text{CO}_2}$ $\newcommand{\SO}{\text{SO}_2}$ $\newcommand{\dist}{\text{dist}}$ $\newcommand{\M}{\mathcal{M}}$ $\newcommand{\N}{\mathcal{N}}$
Let $\M,\N$ be two-dimensional smooth, compact, connected, oriented Riemannian manifolds, with or without boundaries.
Let $f_n \in W^{1,2}(\M,\N)$ with $Jf_n \ge 0$. Suppose that $f_n \rightharpoonup f$ in $W^{1,2}$, and that $\int_{\M} \dist^2( df_n,\CO ) \to0$. Must $df \in \CO$ a.e.?
If the answer is positive, then known regularity results imply that $f$ is smooth, and so $df \in \CO$ everywhere. Thus, it must be either constant or a local diffeomororphism.
Here $\CO =\{\lambda R : R \in \SO\, | \, \lambda \ge 0\} $, where the "copies" of $\SO,\CO$ at each point implicitly depend on the metrics of $\M,\N$ at these points**, so $$2\dist^2(df,\CO)=|df|^2-2Jf=(\sigma_1-\sigma_2)^2,$$ where $\sigma_1, \sigma_2$ are the singular values of $df$.
** For $p \in \M$, $df_p \in \text{Hom}(T_p\M,T_{f(p)}\N)$, and the notion of "$\SO$" depends on the metrics on $T_p\M,T_{f(p)}\N$, and in particular on the image point $f(p)$: $$\text{SO}(T_p\M,T_{f(p)}\N) \subseteq \text{Hom}(T_p\M,T_{f(p)}\N).$$
Here is a proof for the case where $\M=\Omega_1,\N=\Omega_2$ are nice Euclidean domains:
Let $K \subseteq \Omega_1$ be compact. The "higher integrability of Jacobians" implies that $ Jf_n \rightharpoonup Jf $ in $L^1(K)$. Thus, $$ \lim_{n\to \infty} \|df_n\|_{L^2(K)}^2=2\lim_{n\to \infty} \int_K Jf_n=2 \int_K Jf \le\|df\|_{L^2(K)}^2 $$ Since $df_n \rightharpoonup df$ in $L^2$, and the $L^2$-norm is weakly lower semicontinuous, $\|df\|_{L^2(K)}=\lim_{n\to \infty} \|df_n\|_{L^2(K)}$.
In particular, we have $2 \int_K Jf =\|df\|_{L^2(K)}^2$ which implies $f$ is conformal.
The problem with generalizing this argument to manifolds, is that the weak $L^1$ convergence of the Jacobians does no longer hold.
