$\newcommand{\rank}{\operatorname{rank}}$Divide $p-1$ ($p=q^k>2$ with $k>1$, $q$ prime) elements in $\Bbb F_p^\times$ into equal disjoint subsets $S_+,S_-$.
Given square $0-1$ matrix $A$, $\rank_{\Bbb F_p}^+(A)$ is smallest $m$ such that $A=LR$ where $L$ is $n\times m$ matrix, $R$ is non-negative $m\times n$ matrix each containing entries only from $S_+\cup\{0\}$.
Is $\rank_{\Bbb F_p}^+(A)\leq \exp(\log^{\alpha_p}\rank_{\Bbb F_p}(A))$ with a fixed $\alpha_p>0$ over every split into $S_+,S_-$?
My guess is $\rank_{\Bbb F_p}^+(A)\leq\beta_p\rank_{\Bbb F_p}(A)$ holds with a fixed $\beta_p>0$ ($p$ is prime solution below).
Conjecture: In general $\beta_p\leq p^{2k}$ suffices.
Case $p=q^1$ is prime:
$\beta_p\leq p^2$ suffices.
Assume $1\in S_+$. Then each element could be written as sum of atmost $p$ $1$s. This will give worst case bound of $\beta_p\leq p\leq p^2$.
Assume $1\notin S_+$. Represent each element as sum of atmost $p$ $1$s. Pick an element $a$ in $S_+$ with $ba=1$. Write $1$ as sum of $b$ number of $a$s. This will give worst case bound of $\beta_p\leq bp\leq p^2$.