M is complemented in $L_1[0,1]$ iff $L_1/M$ is a $\mathcal{L}_1$-space (Lindenstrauss lifting argument) iff $M^{\perp\perp}$ is complemented in $L_1[0,1]^{**}$ ($X$ $\mathcal{L}_1$-space iff so is $X^{**}$).

The asnwer is no. First, $M$ is not necessarily complemented in $L_1[0,1]$ (Bourgain). In fact, $M$ is complemented in $L_1[0,1]$ iff $L_1/M$ is a $\mathcal{L}_1$-space (Lindenstrauss lifting argument). Second, $L_1/M$ is a $\mathcal{L}_1$-space iff so is its bidual $L_1[0,1]^{**}/M^{\perp\perp}$. Since Lindenstrauss argument applies also to $L_1[0,1]^{**}$, we conclude that $M^{\perp\perp}\equiv M^{**}$ is complemented in $L_1[0,1]^{**}$ iff $M$ is complemented in $L_1[0,1]$.

M is complemented iff L_1/M is a script-L_1 space iff M^{\perp\perp} is complemented in L_1^{**}.

M is complemented in $L_1[0,1]$ iff $L_1/M$ is a $\mathcal{L}_1$-space (Lindenstrauss lifting argument) iff $M^{\perp\perp}$ is complemented in $L_1[0,1]^{**}$ ($X$ $\mathcal{L}_1$-space iff so is $X^{**}$).