I presume you're interested in $c \to \infty$.  Without absolute value you get
$\dfrac{\sqrt{\pi}}{2} c \exp(-c^2/4)$, which is not **quite** of order $\exp(-c^2/4)$.  With absolute value, a lower bound is 
$$ \int_0^\infty \exp(-(x/c)^2) \cos^2 x\; dx = \sqrt{\pi} c (1 + \exp(-c^2))/4$$
so it is not even $O(1)$.  An upper bound is
$$ \int_0^\infty \exp(-(x/c)^2)\; dx = \sqrt{\pi} c/2 $$

EDIT: For  sharp  bounds, consider that for $(n-1/2)\pi \le x \le (n+1/2)\pi$,

$$ e^{-((n-\frac12)\pi/c)^2} |\cos(x)| \ge e^{-(x/c)^2} |\cos(x)| \ge
e^{-((n+\frac12)\pi/c)^2} |\cos(x)|$$
so an upper bound is

$$1 + 2 \sum_{n=1}^\infty e^{-(n-1/2)^2 \pi^2/c^2} = 1 + \theta_2(0,e^{-\pi^2/c^2}) $$
where $\theta_2$ is the second Jacobi theta function,
and a lower bound is
$$  e^{-\pi^2/(4c^2)} + 2 \sum_{n=1}^\infty  e^{-(n+1/2)^2 \pi^2/c^2}
= - e^{-\pi^2/(4 c^2)} + \theta_2(0, e^{-\pi^2/c^2})$$
Using the Jacobi identities (or the Poisson summation formula)
$$\theta_2(0,e^{-\pi^2/c^2}) = \dfrac{c}{\sqrt{\pi}} \left(1 + 2 \sum_{n=1}^\infty (-1)^n e^{-n^2 c^2}\right)$$
In particular, the integral is $1 + c/\sqrt{\pi} + O(c^{-2})$.