Not only a covering, but every piecewise linear map $f$ between finite graphs can be realized in $\Bbb R^3$ in your sense. To see this, pick a triangulation $T$ of the mapping cylinder of $f$ that projects simplicially to some triangulation of $I=[0,1]$. Then pick a generic level-preserving map $g$ of the mapping cylinder in $\Bbb R^3\times I$ that is linear on each simplex of $T$. (Generic here means that the images of any $k+1$ vertices span a $k$-dimensional affine subspace in $\Bbb R^4$ as long as either $k\le 3$, or $k\le 4$ and not all of the $k+1$ vertices have the same $I$-coordinate.)

Then $g$ has only finitely many double points, all in $\Bbb R^3\times (0,1)$. (Say, if a pair of 2-simplices with at most one vertex in common have a 1-dimensional intersection, then their 5 or 6 vertices aren't affinely independent.) If $s$ is the minimal $I$-coordinate of a double point, then the part of $g$ that lands in $\Bbb R^3\times [0,s/2]$ yields the desired homotopy.

I'm quite amazed that this rather standard argument didn't occur to anyone having seen this question in 3 months. Maybe a tag like "geometric topology" could have helped. There is some literature on realizing maps in your sense, which goes under "isotopic realization".