If I understand correctly, you're asking if a nonlinear system of PDE's is involutive, then is the corresponding linearized system also involutive? The answer is yes, but I'm not sure that it's obvious. It is true, as you say, that the symbol, which is the same for both systems, is involutive. However, involutivity also requires conditions on the lower order terms, and these need to be checked. I believe (but am not sure) that this is demonstrated in my monograph "Involutive Hyperbolic Differential Systems" for quasilinear systems.
ADDED: Here's an argument that the homogeneous linearized equation is involutive: Let $$ \Phi(x, u, \partial u) = 0 $$ be a first order involutive system of PDE's. (If the PDE's are higher order, you can always rewrite the system as a first order one)
Given an involutive symbol, you can count the dimension of the space of formal second order solutions there are. In terms of Cartan characters $s_1, \dots, s_n$ (where $n$ is the number of independent variables), it's given by $s_1 + 2s_2 + \cdots ns_n$.
Since the homogeneous linearized equation has the same symbol, it is involutive, if it has the same dimension of formal second order solutions (which is now a linear subspace of the space of all second order Taylor series for the unknown function).
Given any formal $1$-parameter family $u_t$ of 2nd order solutions to the nonlinear system, you have $$ \Phi(x, u_t, \partial u_t) = 0. $$ Differentiating this with respect to $t$, you get $$ \Phi'(x, u_0, \partial u_0)\dot{u} = 0. $$ This implies that the dimension of the space of 2nd order solutions $\dot{u}$ to the linearized system is equal to the dimension of the space of 2nd order solutions to the nonlinear system. By Cartan's test for involutivity (I'm sure there's an equivalent condition in the other definitions of involutivity), the homogeneous linearized system is involutive.