I am looking to show that there is some type of finite speed of propagation property for PDEs of the form
$$ u_{tt} -\Delta u + q u = 0 $$ Where we can assume that $q$ is as smooth or integrable as we like.
We know that for the standard wave equation $$w_{tt} - \Delta w = 0$$ that if $w(x,0) = w_t(x,0) = 0$ for all $x\in B(x_0,r)$, then $w = 0$ on the cone $$ K = \{ (x,t)\in \mathbb{R}^n\times\mathbb{R}:0\leq t\leq r, |x-x_0|\leq r-t \} $$ This result is shown on Page 84 of Evans Partial Differential Equations. It was show at (Prove domain of Dependence Inequality for the Wave Equation?) that the local energy estimate holds if the potential term $q$ is positive. It is often mentioned hyperbolic PDEs have finite speed of propagation, but I can't find a source that proves the claim in the general case.
There are proofs of more general problems (see last paragraph of this answer), but for now let's talk about the equation : $$ u_{tt}-\Delta u = q(x)u \quad (t,x) \in [0,T] \times \mathbb{R}^n \tag{1} \label{1} $$ in which $q:\mathbb{R}^n \rightarrow \mathbb{R}$. (this $q$ is $-q$ of your problem but since there's no assumption on sign of $Im(q)$ it doesn't matter, it's just easier for notations in my proof)
As you said the cone (backward in time) at $(t_0,x_0)$ can be defined as : $$ C(t_0,x_0) :=\{ (t,x)\in [0,T] \times \mathbb{R}^n, 0 \leq t\leq t_0, |x-x_0|\leq t_0-t \} $$ also at every section of time $t \in [0,t_0]$ we define $$ B(x_0,t) = \{x \in \mathbb{R}^n , |x-x_0|\leq t\}. $$ with these notations; the statement of theorem becomes:
Note that $B(x_0,t_0)$ is section of $C(t_0,x_0)$ in $t=0$.
Proof:
define $$ e(t):= \int_{B(x_0,t_0-t)} (u^2+|\partial u|^2)dx $$ where $|\partial u|^2 = |u_t|^2+|\nabla u|^2.$ By rewriting $e(t)$ integral in polar coordinate, we have: $$ e(t) = \int_0^{t_0-t}\int_{\partial B(x_0,s)} (u^2+|\partial u|^2)d\sigma ds . $$ then we have: $$ \dot{e}(t) := \frac{d}{dt} e(t) = \int_{B(x_0,t_0-t)} 2(u u_t + u_t u_{tt} + \underbrace{ \nabla u . \nabla u_t}_{I_1} ) dx - \int_{\partial B(x_0,s)} (u^2+|\partial u|^2)d\sigma . \tag{2} \label{2} $$ we know that : $$ div(u_t \nabla u) = \nabla u . \nabla u_t + u_t \Delta u $$ so we can rewrite $I_1$ in $\eqref{2}$ to get the form of $\eqref{1}$ and use the divergence theorem. continuing from $\eqref{2}$: $$ \dot{e}(t) = 2\int_{B(x_0,t_0-t)} u_t (u + u_{tt} - \Delta u) dx + 2 \int_{B(x_0,t_0-t)} div(u_t \nabla u) dx- \int_{\partial B(x_0,s)} (u^2+|\partial u|^2)d\sigma \\ = 2\int_{B(x_0,t_0-t)} u_t (u + u_{tt} - \Delta u) dx + 2 \int_{\partial B(x_0,s)} u_t \nabla u . \nu d \sigma - \int_{\partial B(x_0,s)} (u^2+|\partial u|^2)d\sigma $$ where $\nu$ is outward unit normal of $\partial B(x_0,s)$. we can bound this integral by $$ 2 |u_t \nabla u . \nu| \leq 2 |u_t| |\nabla u| \leq |u_t|^2 + |\nabla u|^2 = |\partial u|^2 . $$ so we have $$ \dot{e}(t) = 2\int_{B(x_0,t_0-t)} u_t (u + u_{tt} - \Delta u) dx + \int_{\partial B(x_0,s)} \underbrace{\big( (2 u_t \nabla u . \nu - |\partial u|^2) - u^2 \big)}_{\leq 0} d\sigma $$ hence : $$ \dot{e}(t) \leq 2\int_{B(x_0,t_0-t)} u_t (u + \underbrace {u_{tt} - \Delta u}_{qu}) dx = 2\int_{B(x_0,t_0-t)} u_t (1+q(x))u \, dx. \tag{3}\label{3} $$ Let's define $F(r,p):= (1+q(r))p$ . since $F(x,0)=0$ we have $$ (1+q)u-0 = F(x,u)-F(x,0) = \int_0^1 \frac{\partial}{\partial s} F(x,su)ds \\ = \int_0^1 \big( \frac{\partial}{\partial p} F(x,su) \big) \, u \, ds \leq |u| \int_0^1 | \frac{\partial}{\partial p} F(x,su)| ds . $$ therefore $$ |F(x,u)| \leq C |u| $$ where $$ C := \max_{(t,x) \in C(t_0,x_0)} \int_0^1 | \frac{\partial}{\partial p} F(x,s\, u(t,x))| ds. \tag{4} \label{4} $$ Using this inequality in $\eqref{3}$ we get: $$ \dot{e}(t) \leq 2\int_{B(x_0,t_0-t)} u_t F(x,u) \, dx \leq 2C \int_{B(x_0,t_0-t)} |u_t| |u| dx \\ \leq C \int_{B(x_0,t_0-t)} (|u|^2+|u_t|^2) dx \leq C \int_{B(x_0,t_0-t)} (|u|^2+|\partial u|^2) dx = C e(t). $$ $\forall t \in [0,t_0]$ we have $\dot{e}(t) \leq C e(t)$ and also by assumption of theorem $e(0)=0$. using Gronwall's inequality we deduce that $e(t)=0$ for $t \in [0,t_0]$ therefore $u=0$ in $C(t_0,x_0)$ .$\square$
it's worth mentioning that if you look at this proof you can see it's valid for more general form of $F(r,p)$! even form of $F(x,t,u,\partial u,\partial^2 u)$ with some assumptions! so if you look at $\eqref{4}$ you can be more precise when $F(r,p)=(1+q(r))p$ but for abovementioned reason, I didn't simplify it more.