Yes, the principal eigenfunction is strictly positive in $U$ even if the coefficients are only $L^\infty$.
The proof I'm aware of involves quite some machinery from functional analysis. It roughly goes as follows:
(1) First one shows that, for all sufficiently large real numbers $\lambda$, the resolvent $(\lambda + L)^{-1}$, as an operator on $L^2(U)$, maps non-zero functions $f \ge 0$ to functions $u$ that satisfy $u(x) > 0$ for almost all $x \in U$. This follows, for instance, from the Beurling-Deny criterion for bilinear forms.
(2) Spectral theory for positive operators (i.e., variants of the Krein-Rutman theorem) then implies that the first eigenfunction $w_1$ satisfies $w_1(x) > 0$ for almost all $x \in U$.
(3) Now one can use the elliptic Harnack inequality to conclude that even $w_1(x) > 0$ for all $x \in U$.
For the elliptic Harnack inequality in case of measurable coefficients, see for instance "Gilbarg and Trudinger: Elliptic partial differential equations of second order (2001)", Theorem 8.20.
(3') As an alternative to the Harnack inequality employed in step (3), one can use even more functional analysis and operator theory; this is done in Section 4.1 of this paper (link to arXiv). [Disclosure: I am one of the authors of this paper.]
Note: The assumptions needed for the above argument are even a bit weaker then those proposed in the question. For instance, one can allow for lower order terms. This is also contained in [op. cit.].
Best Answer
Not much in general. Consider $n=4$ and
$$u(x) = x_1^2 + x_2^2 - x_3^2 - x_4^2.$$
Then $\text{det}(\nabla^2 u)=1$ but $u$ does not have a sign on, say, the unit ball.
The convex Monge-Ampere equation satisfies a comparison principle. That is, when you have convex sub- and supersolutions $u$ and $v$, and $u\leq v$ on $\partial \Omega$, then $u\leq v$ on $\Omega$. This actually holds in the weaker viscosity sense (see https://arxiv.org/abs/math/9207212). The proof follows a maximum principle argument for nonlinear PDEs.
It's crucially important that the solutions are convex though. In your case, if $u$ is convex, then its maximum occurs on the boundary and so $u\leq 0$ on $\Omega$ (without using Monge-Ampere).