Tubular neighborhoods of embedded manifolds

I suggest to look at the very clear proof of the tubular neighborhood theorem in

Lee, John M., Introduction to smooth manifolds, Graduate Texts in Mathematics. 218. New York, NY: Springer. xvii, 628 p. (2002). ZBL1030.53001., pag 253, I think it's exactly what you need.

He proves the existence of a tubular neighborhood for any embedded submanifold of $\mathbb{R}^n$ (but indeed the proof is the same for any embedded submanifold of a smooth manifold, it just makes use of an auxiliary Riemannian structure to define "normal lines"). Notice that his construction works also in the non-compact case, but in this case the tubular neighborhood has non-constant radius.

If you want $\varepsilon$ to be constant, then looking into the proof you just need a lower bound to the injectivity radius from the submanifold, which you always have if the submanifold is compact (and vice-versa, if you have a tubular neighborhood with constant $\epsilon$, then the injectivity radius is $\geq \varepsilon$.

In any way, curvature has no relation whatsoever with existence of tubular neighborhoods (more precisely, you cannot give sufficient conditions for existence of a uniform tubular neighborhood in terms of curvature bounds, I'm general). Your problem is deeply related with the regularity of the distance from the submanifold. In fact, you can easily build the tubular neighborhoods using the gradient flow of the distance function from the submanifold, which is as smooth as the submanifold in a neighborhood of the latter. To this regard, I can also suggest you to give a look at this paper : Foote, Robert L., Regularity of the distance function, Proc. Am. Math. Soc. 92, 153-155 (1984). ZBL0528.53005.

Ah, and indeed, the magnificent

Gray, Alfred, Tubes, Redwood City, CA etc.: Addison-Wesley Publishing Company. xii, 283 p. (1990). ZBL0692.53001.

You are right that a tubular $\varepsilon$-neighborhood exists, if $M$ is compact. Local existence follows from the properties of the exponential map, global existence by compactness argument, see for example O'Neill, Semi-Riemannian geometry, Proposition 7.26.

Boundedness of principal curvatures is indeed necessary for the existence of a tubular $\varepsilon$-neighborhood, but not sufficient. Consider, for example, a spiral in $\mathbb{R}^3$ with the distance between coils tending to $0$: $$ x(t) = \cos t, \quad y(t) = \sin t, \quad z(t) = \operatorname{arsinh} t. $$

If you don't need $\varepsilon$ to be constant, but allow it to tend to $0$, then a tubular neighborhood exists, this is again O'Neill, Proposition 7.26.

EDIT: The book of John M. Lee cited by Raziel in another answer to this question gives a very clear account on tubes around submanifolds of $\mathbb{R}^n$. O'Neill deals with submanifolds of Riemannian manifolds.

I think you will find your answer in the book "Normally hyperbolic invariant manifolds — the noncompact case," of Jaap Eldering, freely available here. In particular have a look at Theorem 2.33 and the surrounding discussion.