Let $\varepsilon\in(0,1)$ and $c>0$. Note that $$\varphi(x):=e^{-(cx)^2}<\varepsilon\Leftrightarrow|x|>\frac{\sqrt{-\ln\varepsilon}}c\tag1$$ for all $x\in\mathbb R$. Now let $h\in(-1,1)$.
Can we determine a condition (a bound) on $h$ such that whenever this condition is satisfied, then $\varphi(h-k)<\varepsilon$ for all $k\in\mathbb Z$? Or will there always be a $k\in\mathbb Z$ such that $\varphi(h-k)\ge\varepsilon$?
By $(1)$, $$\varphi(h-k)<\varepsilon\Leftrightarrow|h-k|>\frac{\sqrt{-\ln\varepsilon}}c\tag2$$ for all $k\in\mathbb Z$, but this is not helpful. Maybe we can use the reverse triangle inequality, which yields $$|h-k|\ge||h|-|k||=|k|-|h|\tag3$$ for all $k\in\mathbb Z\setminus\{0\}$. Now, for $k\in\mathbb Z$, $$|k|-|h|>\frac{\sqrt{-\ln\varepsilon}}c\Leftrightarrow|h|<|k|-\frac{\sqrt{-\ln\varepsilon}}c\tag4.$$
EDIT
From $(4)$ we should be able to conclude that (assuming $-\ln\varepsilon\le c^2$) if $$|h|<1-\frac{\sqrt{-\ln\varepsilon}}c,\tag5$$ then all $k\in\mathbb Z\setminus\{0\}$ satisfy $(2)$. On the other hand, if $$|h|>\frac{\sqrt{-\ln\varepsilon}}c\tag,6$$ then $(2)$ is satisfied for $k=0$. In conclusion, a possible condition should be $$\frac{\sqrt{-\ln\varepsilon}}c<|h|<1-\frac{\sqrt{-\ln\varepsilon}}c\tag7.$$ Or did I made any mistake?
EDIT 2
I did not, but (obviously) no such $h$ exists (since the left-hand side is at least 1 and the right-hand side is smaller $1$). So, this approach is not helpful.
Your requirement is equivalent to $$[h-a,h+a]\cap\Bbb Z=\varnothing,\text{ where }a=\frac{\sqrt{-\ln\varepsilon}}c,$$ i.e. $$\exists n\in\Bbb Z\quad n+a<h<n+1-a.$$