Can etale $X$-schemes be lifted to $Y$, where $X$ is closed in $Y$?

For a closed embedding (of varieties) $X\to Y$ let $U/X$ be etale. Is is true that there necessarily exists an etale $U'/Y$ such that $U'_X=U$? If this is wrong in general, are there any assumptions that make the statement correct ('my' $X$ is smooth, for example) or any ways to 'repair' it (in particular, I am interested in simplicial schemes).

I need this statement in order to prove my 'conjecture' Does one need to sheafify when defining the inverse image of a sheaf with respect to an embedding? The 'topological' case of my current question is trivial (for manifolds or metric spaces, at least), yet I do not understand what happens in the \'etale setting at all. Certainly, I would also be deeply grateful for any response to my previous question.:)

Upd. Now I also suspect that I need etale tubular neighbourhoods here. Yet any alternative suggestions could be very welcome!

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You probably mean "let $U/X$ be étale". –  Matthieu Romagny Sep 17 '11 at 20:57
Mikhail, I'll assume the correction suggested by Matthieu, but then it isn't true. Take $Y=\mathbb{P}^2$ and $X\subset Y$ a smooth elliptic curve. Any etale cover induced from $Y$ is trivial, but the fundmental group of $X$ is large. You probably want some sort of tubular neighbourhood, but getting this in algebraic geometry seems tricky. If you know Marc Levine, you might ask him. I think he has some version of this, but I haven't seriously looked at it. –  Donu Arapura Sep 17 '11 at 21:56
There is also some old work of Cox in the simplicial scheme setting, which maybe worth looking at as well. –  Donu Arapura Sep 17 '11 at 21:57
Such a statement is true Zariski locally. See EGA IV, Proposition 18.1.1. –  Jonathan Wise Sep 18 '11 at 2:55
@Donu: Note that the extension is not required to be a finite étale map. –  Torsten Ekedahl Sep 18 '11 at 5:06