The answer is yes, and the proof follows easily from the following three observations:
1) If either of $X$ or $Y$ is reflexive, then any bounded operator from $X$ to $Y$ is weakly compact. This follows from the fact that a Banach space is reflexive if and only if its closed unit ball is weakly compact.
2) If $T:L_p[0,1]\rightarrow L_p[0,1]$, $1<p<\infty$, is bounded and if $T$ maps into $C[0,1]$ (note some care in interpretation is needed here since elements of $L_p$ are equivalence classes of functions), then $T$ is bounded when regarded as a map into $C[0,1]$. This follows from Davide Giraudo's argument in the comments, or from an easy application of the Closed Graph Theorem, as suggested by Shalop.
3) $C[0,1]$ has the Dunford-Pettis property: any weakly compact operator from $C[0,1]$ into any Banach space maps weakly convergent sequences to norm convergent sequences. A proof can be found in, e.g., VI.7.4 of Linear Operators, vol. 1, Dunford and Schwartz, or in chapter 5 of Albiac and Kalton's Topics in Banach Space Theory.
So, write your operator as $T=I\circ T_C$ where $T_C:L_p[0,1]\rightarrow C[0,1]$ is defined by $T_C f=Tf$ and $I$ is the "identity" from $C[0,1]$ to $L_p[0,1]$. Both $T_C$ and $I$ are linear, $T_C$ is bounded by 2), and $I$ is bounded since $\Vert\cdot\Vert_p\le\Vert\cdot\Vert_\infty$.
Take a bounded sequence $(x_i)$ in $L_p[0,1]$. It has a weakly convergent subsequence $(y_i)$. Since bounded operators are weak-weak continuous, the image of $(y_i)$ under $T_C$ is weakly convergent. Now, according to 3), $I$, being weakly compact by 1), maps $(T_C y_i)$ to a norm convergent sequence $(Ty_i)$, as desired.
Remark: The Dunford Pettis result is a big gun; I wonder if a more elementary argument can be made.