Let $X$ be a Banach space and let $p\in (1,\infty)$. If $q$ denotes the conjugate exponent to $p$, then $L_q(X^*)$ is easily seen to be isometric to a subspace of $(L_p(X))^*$ via the map $$f\mapsto \int\limits_{[0,1]}\langle \cdot(\omega), f(\omega)\rangle\,{\rm d}\omega\quad (f\in L_q(X^*)).$$ One of the numerous characterisations of the Radon–Nikodym property asserts that $X^*$ has this property with respect to $[0,1]$ endowed with the Lebesgue measure if and only if this map is surjective.

Suppose that $X^*$ fails the Radon–Nikodym property with respect to the Lebesgue measure. Is the range of the above-mentioned map complemented?

My motivation follows from this question of M. González; if the answer were positive, then we would have $\ell_2(X)\not\cong L_2(X)$ for every infinite-dimensional space $X$ with $X^*\cong L_1(\mu)$ and for every infinite-dimensional C*-algebra.

Edit (26.01.2016): This is indeed a known result. It follows from Theorem 9 in:

Z. Hu and B.-L. Lin, Extremal structure of the unit ball of $L_p(\mu, X)^*$,

J. Math. Anal. Appl.200(1996), 567–590.