The answer is ``yes". Slightly more generally, if $(f_n)$ is an orthonormal sequence in $L_2:= L_2(0,1)$ whose $L_1$ norms are bounded away from zero, then there is a subsequence that spans a strongly embedded subspace. It is equivalent (by extrapolation) to get a subsequence that in the $L_p$ norm with $p:= 3/2$ is equivalent to an orthonormal basis in a Hilbert space. The upper $\ell_2$-estimate in $L_p$ for linear combinations of the $f_n$ comes for free because $(f_n)$ is orthonormal and $p<2$. For the lower estimate, use the fact that $L_p$ has an unconditional basis to pass to a subsequence of $(f_n)$ that is unconditional. By the cotype $2$ property of $L_p$ when $p<2$, linear combinations of that subsequence of $(f_n)$ has a lower $\ell_2$ estimate.

EDIT Oct. 21, 2019: This question is closely related to the Maurey--Rosenthal [MR] problem whether there is a normalized weakly null sequence in $L_1$ that has no unconditionally basic subsequence. An example was given in [JMS]. It is natural to ask whether such a sequence can be bounded in $L_p$. The simple answer above says ``no" if $p\ge 2$. However, there are examples that are bounded in $L_p$ for all $p<2$--see the conditions (7) and (8) in [JMS].

[JMS[ Johnson, William B.; Maurey, Bernard; Schechtman, Gideon Weakly null sequences in L1. J. Amer. Math. Soc. 20 (2007), no. 1, 25–36.

[MR] Maurey, B.; Rosenthal, H. P.
Normalized weakly null sequence with no unconditional subsequence.
Studia Math. 61 (1977), no. 1, 77–98.