No, you can't always find two such matrices.

Let $n \gt 2$. Choose a $2$-dimensional subspace $V$ of $\mathbb{Q}^n$ that does not intersect the positive orthant except at $\vec 0$. Let $M$ be a matrix whose columns span $V$. Any decomposition of $M$ into $M_+ - M_-$, with each of these of rank $1$ and in $\mathbb{Q}^{n\times n}_{\ge 0}$ would produce a basis for $V$ consisting of vectors from the positive orthant, which by construction was impossible. In fact, we can't decompose $M$ as a sum of rank-$1$ matrices if we require even one of the summands to be all-nonnegative.

Example: $ \begin{pmatrix} -2 & 1 & 1 \newline 1 & -2 & 1 \newline 1 & 1 & -2 \end{pmatrix}$ has a column space consisting of vectors of sum $0$, which does not intersect the positive orthant except at $\vec{0}$ so it can't be decomposed this way.

No, you can't always find two such matrices.

Let $n \gt 2$. Choose a $2$-dimensional subspace $V$ of $\mathbb{Q}^n$ that does not intersect the positive orthant except at $\vec 0$. Let $M$ be a matrix whose columns span $V$. Any decomposition of $M$ into $M_+ - M_-$, with each of these of rank $1$ and in $\mathbb{Q}^{n\times n}_{\ge 0}$ would produce a basis for $V$ consisting of vectors from the positive orthant, which by construction was impossible. In fact, we can't decompose $M$ as a difference of rank-$1$ matrices if we require even one of them to be all-nonnegative.

Example: $ \begin{pmatrix} -2 & 1 & 1 \newline 1 & -2 & 1 \newline 1 & 1 & -2 \end{pmatrix}$ has a column space consisting of vectors of sum $0$, which does not intersect the positive orthant except at $\vec{0}$ so it can't be decomposed this way.