Let $f:\Sigma \to (M,g)$ be an immersed surface in a Riemannian manifold with normal bundle $N$. The second fundamental form $A$ of $f(\Sigma)$ is a section of $S^2(T^*\Sigma)\otimes N$, the bundle of real bilinear symmetric forms with values in $N$.
Suppose that $TM|_{f(\Sigma)}$ admits an almost complex structure (e.g. $M$ is almost complex). Then we can decompose $S^2(T^*\Sigma)\otimes N$ into three parts: $$ (S^{2,0}(T^*\Sigma)\otimes_{\mathbb{C}} N ) \oplus (S^{1,1}(T^*\Sigma)\otimes_{\mathbb{C}} N ) \oplus (S^{0,2}(T^*\Sigma)\otimes_{\mathbb{C}} N ). $$ The first summand is complex bilinear, the second is the symmetrization of complex linear in the first argument and antilinear in the second argument. The third summand is anti bilinear.
Under this decomposition, the second fundamental form $A$ splits into three parts. If $J$ is the almost complex structure on $TM|_{f(\Sigma)}$, then \begin{align} A^{2,0}(x,y)&=A(x,y)-JA(Jx,y)-JA(x,Jy)-A(Jx,Jy) \\ A^{1,1}(x,y)&=A(x,y)+A(Jx,Jy) \\ A^{0,2}(x,y)&=A(x,y)+JA(Jx,y)+JA(x,Jy)-A(Jx,Jy). \end{align}
Note that if $e_1,e_2=Je_1$ is an orthonormal basis for $T_p\Sigma$, then $A^{1,1}(e_1,e_1)$ is the mean curvature of $f(\Sigma)$. Thus $A^{1,1}\equiv 0$ iff $f(\Sigma)$ is a minimal surface. Also, the $A^{2,0}$ and $A^{0,2}$ components are conformally invariant.
Question: I am interested in what the condition $A^{0,2}\equiv 0$ says about $f(\Sigma)$. Given $(M,g)$, does there always exist a nonconstant immersion $f:S^2\to M$ with $A^{0,2}\equiv 0$?
Note that any totally umbilic immersion has $A^{0,2}\equiv 0$, as can easily be checked on a basis. Any references would be appreciated.