If $\lim_{x \to \pm \infty} g(x) = 0$ then the assertion is false. Define $h(x) := g(-x)$, $H(\alpha) := \int_{-\infty}^\infty \frac{1}{\frac{h(x+\alpha)}{h(x)}+k} ~ h(x) ~ dx = \int_{-\infty}^\infty \frac{1}{\frac{g(-x-\alpha)}{g(-x)}+k} ~ g(-x) ~ dx = \int_{-\infty}^\infty \frac{1}{\frac{g(x-\alpha)}{g(x)}+k} ~ g(x) ~ dx$. (We only need this since $\alpha \geq 0$.) Then again $h$ is a density and by Lebesgue's theorem on dominated convergence $\lim_{\alpha \to \infty} F(\alpha) = \frac{1}{k} = \lim_{\alpha \to \infty} H(\alpha)$. Define $F(\alpha) := H(-\alpha)$ for $\alpha <= 0$. If the assertion is always true then $\alpha \to F(\alpha)$ is always decreasing for $\alpha \leq 0$ and increasing for $\alpha \geq 0$.

But this then also holds true for $F_\beta(\alpha) := \int_{-\infty}^\infty \frac{1}{\frac{g(x+\beta-\alpha)}{g(x+\beta)}+k} ~ g(x+\beta) ~ dx = F(\alpha-\beta)$, $\beta \in \mathbb{R}$. As a consequence any $\beta \in \mathbb{R}$ is a minimum point of $F$, i.e. $F \equiv \frac{1}{k}$. This is only possible if $g \equiv 0$, a contradiction.

This sort of reasoning of course holds for unimodal densities too.

Edit: This answer is not correct. Actually $F_\beta(\alpha) = F(\alpha)$.