Question. Suppose that $u_1, u_2\in C^2(\mathbb R^{d+1})$ are real-valued, satisfy $\partial_t^2 u = \Delta u$, and $$|u_1(t, x)|=|u_2(t, x)|, \qquad \forall (t, x)\in \mathbb R^{d+1}.$$ Does it follow that either $u_1=u_2$ or $u_1=-u_2$?
This property holds for real-analytic functions. Indeed, either $u_1=u_2=0$ identically (trivial case), or there is $z_0=(t_0, x_0)$ such that $u_1(z_0)\ne 0$. Now, if $u_1(z_0)=u_2(z_0)$, then, by continuity, $u_1=u_2$ on a neighborhood of $z_0$, and so they coincide everywhere, because they are analytic. Otherwise, by the same argument, $u_1=-u_2$.
However, solutions to the wave equation need not be real-analytic. Instead, they satisfy the finite speed of propagation; if $u_1= u_2$ or $u_1=-u_2$ on the cylinder $\{(t, x)\ |\ |t-t_0|<a, |x-x_0|<a\}$, then the same is true on the cone $$ \{|t-t_0|<2a, |x-x_0|<2a-|t-t_0|\}.$$ This implies that the property holds if the zero set $Z=\{u_1=u_2=0\}$ is a space-like hypersurface. Indeed, by assumption it must be either $u_1=u_2$ or $u_1=-u_2$ on either side of $Z$, but there are cones that cross $Z$, so it must be $u_1=u_2$ or $u_1=-u_2$ everywhere; see picture. However, this argument cannot be applied if $Z$ is a time-like or light-like hypersurface.
The answer appears to be negative. The functions $$ u_1(t, x)=(t-x)^3, \qquad u_2(t, x)=|t-x|^3$$ satisfy the wave equation $\partial_t^2 u=\partial_x^2$ on $\mathbb R^{1+1}$, belong to $C^2(\mathbb R^{1+1})$, satisfy $|u_1(t, x)|=|u_2(t, x)|$ at all $(t, x)\in \mathbb R^{1+1}$. However, it is not true that $u_1=u_2$ or $u_1=-u_2$.
Remarks. The function $u_2$ is not real-analytic, and the zero-set of $u_1$ and $u_2$ is light-like. We expected that; see the OP.
More examples can be generated by taking $u_1=f(x-t)$ and $u_2=|f(x-t)|$ for suitable functions $f\colon \mathbb R\to \mathbb R$; the only requirement on $f$ is that it is odd and that it vanishes at $0$ sufficiently rapidly, so that $|f|$ is differentiable. This way, we can produce examples in $C^\infty$ with initial data of compact support.