As a partial answer, there is no smooth partition of $\mathbb{R}^3$ into smooth loops.  For, consider the quotient space; it is necessarily a [two-orbifold][1] (locally a surface, possibly having isolated points each having a neighbourhood modelled on $\mathbb{R}^2$ modulo a finite-order rotation). 

Since $\mathbb{R}^3$ is non-compact and simply connected, so is the quotient.  The only such orbifold is $\mathbb{R}^2$.  Thus the partition gives $\mathbb{R}^3$ the structure of an $S^1$-bundle over $\mathbb{R}^2$.  

Now, $\mathbb{R}^2$ is contractible.  So any $S^1$-bundle over it is isomorphic to $S^1 \times \mathbb{R}^2$.  This has fundamental group $\mathbb{Z}$ and we are done.

<hr>

Smoothness is not really necessary.  Instead we need that every loop has a small neighbourhood which is homeomorphic to a "standard fibering of a solid torus".  If this holds, then the partition gives $\mathbb{R}^3$ the structure of a [Seifert fibered space][2].  The [non-compact case][3] is more difficult than the [compact case][4], but there is still a rich theory. 

The extra structure (of namely having good neighbourhoods) holds when the parts of the partition are (continuously varying) round circles.  So the answer to Question 1 is no. Question 2 feels more subtle; after all, continuous loops in space can be pretty badly behaved! So obtaining the bundle structure seems more difficult. 


  [1]: https://en.wikipedia.org/wiki/Orbifold
  [2]: https://en.wikipedia.org/wiki/Seifert_fiber_space
  [3]: https://mathoverflow.net/questions/462959/non-compact-seifert-manifolds
  [4]: https://pi.math.cornell.edu/~hatcher/3M/3M.pdf