This is still true, although as Francesco says in his comment above, it is vacuously so in general. 

Proof : if $f:(V,J)\to\mathbb{C}$ is such a function, namely $df\circ J=i\\,df$, then (obviously) $d(df\circ J)=0$.

But the second order operator $f\mapsto (d(df\circ J))^{1,1}$ from functions to  $(1,1)$-forms has the "same" principal symbol at each point as in the integrable case (the "plurisubharmonic Hessian", so to speak). 

In particular you can compose it with contraction by a positive smooth $(1,1)$ form (given by any hermitian metric) to obtain a Laplace operator, which satisfies the maximum principle.