Since $\cos(-x) = \cos(x)$ we can reduce the integration range to $(0,\pi)$, and then do the change of variable $u = \cos(x)$: $$ \int\limits_{-\pi}^\pi \frac{\mathrm{d} x}{\sqrt{(t-2 \cos(x))^2-4}} = \int_0^\pi \frac{\mathrm{d} x}{\sqrt{\left(\frac{t}{2} - \cos(x)\right)^2 -1}} = \int_{-1}^1 \frac{2 \mathrm{d} u}{\sqrt{1-u^2} \sqrt{4 u^2 + 4 u t + t^2-4}} $$ Now perform a change of variables: $$ u = \frac{v-z}{1- v z}, \qquad \text{where} \quad v = \frac{t - \sqrt{t^2-16}}{4} $$ that maps $-1 into $-1, that leads to $$ \int_{-1}^1 \frac{ v \, \mathrm{d} z}{\sqrt{1-z^2} \sqrt{1-v^4 z^2}} = 2 v \cdot \mathrm{K}\left( v^4\right) $$ This is not exactly what the OP asked for, but quite elegant nonetheless.
Here is a confirmation of the equivalence in Mathematica:

Added:
The equivalence of the above answer to the one conjectured by OP and established by @J.M. is through the
quadratic transformation: $$ \mathrm{K}(z) = \frac{2}{1+\sqrt{1-z}} \mathrm{K} \left( \left(\frac{1-\sqrt{1-z}}{1+\sqrt{1-z}} \right)^2 \right) $$ where
$z = \frac{16}{t^2}$. Indeed: $$ \frac{2}{1+\sqrt{1-\frac{16}{t^2}}} = \frac{2t}{t + \sqrt{t^2-16}} = \frac{2t}{t + \sqrt{t^2-16}} \cdot \frac{t - \sqrt{t^2-16}}{t - \sqrt{t^2-16}} = \frac{t}{4} \cdot \frac{t-\sqrt{t^2-16}}{2} $$ and similarly: $$ \left(\frac{1-\sqrt{1-\frac{16}{t^2}}}{1+\sqrt{1-\frac{16}{t^2}}} \right)^2 = \left(\frac{t-\sqrt{t^2-16}}{t+\sqrt{t^2-16}} \cdot \color\green{ \frac{t-\sqrt{t^2-16}}{t-\sqrt{t^2-16}} } \right)^2 = \left(\frac{\left(t-\sqrt{t^2-16}\right)^2}{16} \right)^2 $$ Combining, we arrive at the equality: $$ \frac{4}{t} \mathrm{K}\left(\frac{16}{t^2}\right) = \frac{t-\sqrt{t^2-16}}{2} \mathrm{K}\left(\left( \frac{t-\sqrt{t^2-16}}{4} \right)^4\right) $$