It is not true. Consider a continuous linear surjection of $\ell_1$ onto $c_0$; such a map exists and is strictly singular (every operator from $\ell_1$ to $c_0$ is strictly singular).
What is true is the following result Pelczynski:
Theorem. Let $\mu$ be a non-trivial measure on the field of all Borel subsets of some topological space and $T: X\longrightarrow L_1(\mu)$ a bounded linear operator. The following are equivalent:
- $T$ is not weakly compact.
- $T$ factors the identity operator of $\ell_1$.
- There exists a complemented subspace $Y$ of $X$ such that $T|_Y$ is an isomorphism, $Y$ (and hence $T(Y)$ also) is isomorphic to $\ell_1$ and $T(Y)$ is complemented in $L_1(\mu)$.
- $T$ is strictly cosingular.
Moreover, if $X$ has the Dunford-Pettis property, then 1.-4. above are equivalent to:
$5.$ $T$ is strictly singular.
This result is proved in Pelczynski's paper On strictly singular and strictly cosingular operators. II. Strictly singular and strictly cosingular operators in $L_1(\nu)$ spaces, Bull. Acad. Polon. Sci. Ser. Sci. Math. Astronom. Phys. 13 (1965) 37-41. It generalises earlier joint work of Kadets and Pelczynski that showed that a nonreflexive subspace of an $L_1(\mu)$ space contains a subspace that is isomorphic to $\ell_1$ and complemented in the ambient space $L_1(\mu)$.
Pelczynski's theorem above for non-weakly compact operators into $L_1(\mu)$ spaces is, in a sense, dual to his theorem asserting that an operator from a $C(K)$ space is non-weakly compact if and only if it fixes a copy of $c_0$. As Pelczynski points out in the beginning to part I. of the above paper, "these results are closely connected with criteria of weak compactness of linear operators in $C(S)$ and $L_1(\nu)$ spaces due to Grothendieck".