The statement is true and let us prove it by induction on $n$.
Denote $\mathbb{Z}/2\mathbb{Z}:=\mathbb{Z}_2:=\{0,1\}$. When $n=1$, the statement is clearly true. Now assume that when $n\le n_0$, the statement is true for every $m\ge 1$, and let us show the statement is true for $n:=n_0+1$ and every $m\ge 1$.
Consider $(\mathbb{Z}_2)^n$ as $(\mathbb{Z}_2)^{n-1}\times\mathbb{Z}_2$ and denote by $P$ and $Q$ the projections of $(\mathbb{Z}_2)^n$ to its first $n-1$ coordinates and its last coordinate respectively, i.e.
$$P: (\mathbb{Z}_2)^n\mapsto (\mathbb{Z}_2)^{n-1},\quad(a_1,\dots,a_{n-1},a_n)\mapsto (a_1,\dots,a_{n-1}),$$
and
$$Q: (\mathbb{Z}_2)^n\mapsto \mathbb{Z}_2,\quad(a_1,\dots,a_n)\mapsto a_n.$$
Up to a rearrangement of $v_1,\dots,v_m$, we may assume that there exists $0\le m'\le m$, such that $Qv_i\ne 0$ if and only if $m'<i\le m$. If $m'=0$, simply choose $f=Q$ and we are done, so let us assume that $m'\ge 1$. By definition, $Pv_1,\dots,Pv_{m'}\ne 0$. By the induction assumption on $n$, there exists a linear function $g:(\mathbb{Z}_2)^{n-1}\to \mathbb{Z}_2$, such that
$$|\{ 1\le i\le m' \mid g(P v_i) \neq 0\}| \ge \frac{1}{2}m'.\tag{1}$$
For $j=0,1$, define
$$f_j:(\mathbb{Z}_2)^n\to \mathbb{Z}_2,\quad v\mapsto g(Pv)+j\cdot Qv.\tag{2}$$
By definition, for $j=0,1$, $f_j$ is linear and $f_j(v_i)=g(P v_i)$ when $1\le m\le m'$. Then by $(1)$, for both $j=0$ and $j=1$,
$$|\{ 1\le i\le m' \mid f_j(v_i) \neq 0\}| \ge \frac{1}{2}m'.\tag{3}$$
If $m'=m$, then we are done. Otherwise, note that by $(2)$ and the choice of $m'$, for $j=0,1$,
$$f_j(v_i)=0 \iff g(Pv_i)=1-j\iff f_{1-j}(v_i)=1,\quad \forall\, m'<i\le m.$$
It follows that
$$|\{ m'< i\le m \mid f_0(v_i) \neq 0\}|+|\{ m'< i\le m \mid f_1(v_i) \neq 0\}|=m-m'.\tag{4}$$
Combining $(3)$ and $(4)$, we can conclude that the statement is true for either $f=f_0$ or $f=f_1$, which completes the proof.
Remark: Thanks to PeteL.Clark's nice question in the comment, I realized that the conclusion can be generalized to arbitrary finite field with the same argument as above.
Claim: Let $\mathbb{F}_q$ be the finite field of $q$ elements and let $v_1,\dots,v_m$ be nonzero vectors in $(\mathbb{F}_q)^n$. Then
there exists a linear function $f:(\mathbb{F}_q)^n\to\mathbb{F}_q$,
such that $$|\{1\le i\le m\mid f(v_i)\ne 0\}|\ge\frac{(q-1)m}{q}.\tag{5}$$
Sketch of Proof:
Define projections $P:(\mathbb{F}_q)^n\to (\mathbb{F}_q)^{n-1}$, $Q:(\mathbb{F}_q)^n\to\mathbb{F}_q$ similarly. Define $m'$ and rearrange $v_1,\dots,v_m$ similarly. By induction, we may assume that there exists a linear function $g:(\mathbb{F}_q)^{n-1}\to\mathbb{F}_q$, such that for $f_j=g\circ P+j\cdot Q$, $0\le j<q$,
$$|\{1\le i\le m'\mid f_j(v_i)\ne 0\}|\ge\frac{(q-1)m'}{q}.\tag{6}$$
By definition of $m'$ and $f_j$, for every $m'<i\le m$, $f_0(v_i),\dots, f_{q-1}(v_i)$ are pairwise different, so one and only one of them is $0$. It follows that for some $0\le j<q$,
$$|\{m'< i\le m\mid f_j(v_i)\ne 0\}|\ge\frac{(q-1)(m-m')}{q}.\tag{7}$$
$(5)$ follows from $(6)$ and $(7)$ for $f=f_j$.