As asked, the answer to the question is 'no'. The simply-connected cover $f:\mathbb{R}^2\to S$ of Sherck's first surface $S$ (which is defined in $\mathbb{R}^3$ by the equation $\mathrm{e}^x\cos y =\cos x$) has strictly negative curvature and is complete but there do not exist global coordinates $s,t$ on $\mathbb{R}^2$ (the domain of $f$) whose level sets are the asymptotic curves.
What's true is that there are functions $u$ and $v$ on $S$ itself such that $(u,v):S\to\mathbb{R}^2$ is a coordinate chart on $S$ and the level curves of $u$ and of $v$ are (unions of) asymptotic curves on $S$.
However, $S$ is not simply-connected; the image $(u,v)(S)$ in $\mathbb{R}^2$ is the entire plane minus the integer lattice $\mathbb{Z}^2$. It follows from this that, if $f:\mathbb{R}^2\to S$ is the simply-connected cover of $S$, then the functions $u{\circ}f$ and $v{\circ }f$ are not coordinates on $\mathbb{R}^2$, as they do not distinguish points. If there were global asymptotic coordinates $s$ and $t$ on $\mathbb{R}^2$, then (after switching $s$ and $t$ if necessary) they would have to satisfy $\mathrm{d}s\wedge \mathrm{d}(u{\circ}f)=0$ and $\mathrm{d}t\wedge \mathrm{d}(v{\circ}f)=0$, and it is easy to see that no such functions can define a coordinate system and distinguish points.
In the general case, I think it will not be easy to formulate a necessary and sufficient condition for global asymptotic coordinates to exist on the simply-connected cover of an immersed surface of negative curvature in $\mathbb{R}^3$.