These are the unlink.
If you take any $z$ cross-section, you see a discrete collection of points in the plane, say countably many. So you have a map $\varphi: \mathbb{N}\times \mathbb{R} \to \mathbb{R}^3$ such that $\varphi(n, z) = (f_1(n,z),f_2(n,z), z)$. Claim: one may find a smoothly varying family of diffeomorphisms $\phi_z: \mathbb{R}^2\to \mathbb{R}^2$ such that $\phi_z((n,0))=(f_1(n,z),f_2(n,z))$. Then we obtain a diffeomorphism $\Phi: \mathbb{R}^3\to\mathbb{R}^3$ defined by $\Phi(x,y,z)=\phi_z(x,y)$, which sends the infinite unlink $\mathbb{N}\times \{0\}\times \mathbb{R}$ to the given braid. Since $Diff_+(\mathbb{R}^3)$ is connected, this diffeo. may be achieved by ambient isotopy.
Here's a proof of the claim. First, let's see how to extend a single point moving smoothly in the plane to a family of diffeos. Let $\phi^1_z(x,y)=(f_1(0,z)+x,f_2(0,z)+y)$. Then $\phi^1_z(0,0)=(f_1(0,z),f_2(0,z))$. Thus, we can "straighten" a single strand by composing with this diffeomorphism.
Assume we have "straightened" $n$ strands. To straighten the $n+1$st strand, we choose the same diffeo. above tailored to this strand. This diffeo. is tangent to a 1-parameter family of vector fields in the plane. We may modify these vector fields by a bump function to be zero in a neighborhood of the first $n$ points which avoids the $n+1$st point. Then integrate this vector field to get a family of diffeomorphisms which fixes the first $n$ points. One may be a bit more careful with this induction to make sure that as $n\to \infty$, this family of diffeomorphisms is eventually constant on compact subsets of the plane. Then one obtains in the limit the desired 1-parameter family of diffeos. which straightens the link.