It seems that my previous answer was completely wrong, and that the fact is **true** for all nonzero integer (not necessarily prime) $p$.

Let $M=[m_{ij}]_{i\leq n,\;j\leq m}$. If there is no $x$ satisfying the requirements (we regard all entries as integers, not residues!), then one of the sums of the form $\sum_{i\leq n} m_{ij}x_i$, $j\leq m$, is divisible by $p$ for every integer vector $x$. Equivalently,
$$
\prod_{j\leq m}\left(1-\exp\left(\frac{2\pi i}p\sum_{i\leq n} m_{ij}x_i\right)\right)=0
$$
for every integral vector $x$.

After expanding all the brackets, we get the equality of the form
$$
\sum_k c_k\xi_{1k}^{x_1}\dots \xi_{nk}^{x_n}=0,
$$
where $c_i$ are some (integer) numbers, and $\xi_{ik}$ are some nonzero complex numbers; after collecting terms, we may assume that all the $n$-tuples $(\xi_{1k},\dots,\xi_{nk})$ are distinct. Due to the usual Vandermonde argument, this may happen only if $c_k=0$ for all $k$. This means, in particulat, that the term $1^m$ cancels with some other term, which has the form
$$
(-1)^t\prod_{j\in T}\exp\left(\frac{2\pi i}p\sum_{i\leq n} m_{ij}x_i\right)
=(-1)^t\prod_{i\leq n}\left(\exp\left(\frac{2\pi i}p\sum_{j\in T}m_{ij}\right)\right)^{x_i}.
$$
for some $\varnothing\neq T\subseteq \{1,2,\dots,m\}$ with $t=|T|$. This yields that $\sum_{j\in T}m_{ij}$ is divisible by $p$ for all $i\leq n$, so the characteristic column of $T$ (taken as $v$) violates the condition imposed on $M$.

Moreover, one may notice that there is such $T$ with odd cardinality; so we have proved that $v$ may be chosen with odd number of ones in it.

**ADDENDUM.** The `usual Vandermonde argument' I meant consists in the following. Assume that $(\xi_{1,k},\dots,\xi_{n,k})$, $k=1,\dots,K$, are distinct tuples of complex numbers. Then the arrangements $(\xi_{1,k}^{x_1}\cdots\xi_{n,k}^{x_n})_{x_i\in\mathbb Z}$ are linearly independent. The proof goes by the induction on $n$; the base case $n=1$ is classical Vandermonde. For the step of induction, assume the linear dependence, collect the terms with identical $\xi_{1,k}$ and apply first the hypothesis (to prove that one of the coefficients of $\xi_{1,k}^{x_i}$ does not vanish for some choice of $x_2,\dots,x_n$) and then the base to finish the step.