I would like to offer another explanation of the impossibility of foliating $R^3-0$ by tori (or by higher genus closed surfaces), at least in the $C^\infty$ case.

Previously I commented that "foliations are rather far from fibrations". Closer to the truth, foliations are `submersions' onto their (potentially very weird!) leaf spaces. So the distance between foliations and fibrations is, in some respects, comparable with the distance between submersions and fibrations. In the case of smoothly foliating open manifolds by compact submanifolds, the distinction however is very small -- in fact, I want to claim below that foliations in this setting are exactly fibrations.

The starting point is Ehresmann's fibration theorem: if $f:V\to M$ is a proper submersion of smooth manifolds, then $f$ is a locally trivial fibration. A proof can be found in Brocker & Janich's "Introduction to differential topology", section 8.12.

Hence if we have a proper smooth function $f$ on $R^3-0$ having no critical points, then the fibres $f^{-1}(pt)$ foliate $R^3-0$ by compact embedded submanifolds. Ehresmann's theorem tells us $f$ is a locally trivial fibration, and of course as $R$ is contractible, $f$ actually defines a globally trivial fibration. In otherwords, all the fibres are diffeomorphic and we have $R^3-0 \simeq f^{-1}(pt) \times R$. From here we can determine that any fibre must be $\simeq S^2$.

An important point which needs some further justification is this: given a smooth foliation $\mathscr{F}$ of $R^3-0$ by, say, compact tori, how do I know that the quotient map from $R^3-0$ to the leaf space is a smooth submersion of $R^3-0$ onto a smooth $1$-manifold? I see how compactness ensures distinct leafs remain separated (and so the quotient is hausdorff), however i am not clear on pinning down the smooth structure of the quotient map.

Granting the above point, we would know that the quotient is necessarily a noncompact smooth 1-manifold, i.e. the real line $R$, and hence the total space must be a product $T\times R$ -- which we know it isn't.

These arguments, however, suffer the shortcoming of not being able to establish whether or not $R^3-0$ can be foliated by punctured surfaces, a question which seems interesting itself.