**Yes**, $J^*$ is subprojective.

Suppose that $M$ is a closed subspace of $J^*$. By Mazur's theorem, without loss of generality we additionally suppose that $M$ has a normalised basis $(f_k)_{k=1}^\infty$. Let $\iota\colon M\to J^*$ be the inclusion map and let $Q\colon J^{**}\to M^*$ be its (surjective) adjoint. Denote by $(f^*_k)_{k=1}^\infty$ the coordinate functionals of $(f_k)_{k=1}^\infty$.

Choose a bounded sequence $(z_n)_{n=1}^\infty$ in $J^{**}$ such that $Qz_n = f_n^*$ ($n\in \mathbb{N}$). As $J$ is quasi-reflexive, by Rosenthal's $\ell_1$-theorem, $(z_n)_{n=1}^\infty$ contains a weakly Cauchy subsequence, call it $(z^\prime_n)_{n=1}^\infty$. As $(z^\prime_n)_{n=1}^\infty$ fails to have a convergent subsequence, we may suppose that the sequence $(z^\prime_{2n+1}-z^\prime_{2n})_{n=1}^\infty$ is semi-normalised. However, by a result of Andrew [1, Theorem 2.1], every semi-normalised weakly null sequence in $J$ (so also in $J^{**}$ as $J^{**}\cong J$) contains a sequence spanning a complemented copy of $\ell_2$, so $(z^\prime_{2n+1}-z^\prime_{2n})_{n=1}^\infty$ contains such subsequence, say $(w_n)_{n=1}^\infty$. By reflexivity of $\ell_2$, $(w_n)_{n=1}^\infty$ is shrinking so the adjoint of a projection from $J^{**}$ onto $[w_n]_{n=1}^\infty$ is a projection onto a subspace of $M$. As $M\subset J^* \subset J^{***}$, the conclusion follows.

References:

- A. Andrew, Spreading basic sequences and subspaces of James' quasi-reflexive space,
*Math. Scand.* **48** (1981), 276–282.