$Ax=b$ has a solution over $\Bbb{F}_p$ for every prime $p~\implies$ existence of real solution??

This is true. There will exist a rational solution.

Let $V\subseteq \Bbb{Q}^m$ be the column space of $A$. If there is no solution $x\in\Bbb{Q}^n$ this means that the column space of the augmented matrix $A'=(A|b)$ has higher dimension than $r=\dim_{\Bbb{Q}}V$. In other words, $A'$ has rank $r+1$ while $A$ has rank $r$ only.

This means that all the size $r+1$ minors of $A$ vanish, but there exists at least one non-vanishing minor $\Delta$ of $A'$ of the same size. Let $p$ be a prime that is not a factor of $\Delta$. If $\overline{A}$ and $\overline{A'}$ are the above matrices reduced modulo $p$ we see that $\overline{A'}$ has a non-vanishing minor $\overline{\Delta}$ of size $r+1$ whereas all the $(r+1)\times (r+1)$ minors of $A$ are equal to zero. This implies that $\overline{b}$ is not in the $\Bbb{F}_p$-span of the columns of $\overline{A}$. Therefore the system won't have a solution modulo $p$ either.

For the bonus question, the condition implies the determinant is odd, hence not zero, hence the matrix is invertible over the reals (indeed, over the rationals).