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 $ c/\sqrt{\pi} + O(1)$. I expect that further analysis could change this $O(1)$ to something much better.