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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 surface. Since $\mathbb{R}^3$ is non-compact and simply connected, so is the quotient. So the quotient space is homeomorphic to $\mathbb{R}^2$. Thus the partition gives $\mathbb{R}^3$ the structure of an $S^1$-bundle over $\mathbb{R}^2$. But the fundamental group of any such bundle is $\mathbb{Z}$, contradicting the fact that $\mathbb{R}^3$ is simply connected.