A LTI system is described by
$$\dot{x}=Ax+Bu$$
If $A$ is of the form:
$$A=\begin{pmatrix} A_1(x_1,…,x_j)\\ A_2(x_{j+1},…,x_n) \end{pmatrix}$$
The meaning of this structure of $A$ is that only the first $j$ states matter for the first $j$ derivatives and similar for the other part of $A$. $B$ is of the form :
$$B=\begin{pmatrix} B_1\\ B_2 \end{pmatrix}$$
And $u$ can only assume the values $\pm1$, then can we say that the we have two decoupled subsystems and thus we can consider as independent
$$\dot{x}_1=A_1x_1+B_1u \qquad (x_1,\dot{x}_1) \ \text{range from} \ 1 \ \text{to} \ j$$
and
$$\dot{x}_2=A_2x_2+B_2u \qquad (x_1,\dot{x}_2) \ \text{range from} \ j+1 \ \text{to} \ n$$
or, in words, if the states are not related each other but by an exogenous input, can we say the states are independent?
In control theory, systems are considered decoupled with regards to the input-state (or input-output) relation. That is, it must be the case that there is one input that drives the $x_1$ states, and another input that drives the $x_2$ states. So, no, the systems are not decoupled. The (single) independent variable (control) $u,$ determines two independently-varying states $x_1$ and $x_2.$ You need more than one input to perform a decoupling of this input-state relation. The range of allowable $u$ makes no difference to this end.
The LTI system is properly decoupled if the Transfer Function Matrix from the inputs to the outputs are diagonal.