What you need is the following:
Let $v \in C^\infty_c(U)$ and $w\in C^\infty(\bar{U})$, we have
$$ \left(\int_U Dv \cdot Dw ~\mathrm{d}x\right)^2 \leq C \int_U |v|^2 \mathrm{d}x \int_U |D^2 w|^{2}\mathrm{d}x \tag{*}$$
This follows by directly integrating by parts (the boundary terms vanish as $v$ has compact support).
Now, given $u \in H^1_0(U) \cap H^2(U)$, let $v_i \to u$ in $H^1_0$ and $w_i \to u$ in $H^2(U)$ where $v_i \in C^\infty_c(U)$ and $w_i \in C^\infty(\bar{U})$.
By the strong convergence in $H^1_0$ and $H^2$ respectively, we have that for any function $f\in L^2$ we have
$$ \lim_{\ell \to \infty}\int_U \partial_{x^j} v_\ell f \mathrm{d}x = \lim_{\ell \to \infty}\int_{U} \partial_{x_j} (v_\ell - u + u) f \mathrm{d}x = \int_{U} \partial_{x^j} u f \mathrm{d}x + \lim_{\ell\to\infty}\int_{U} (\partial_{x_j}v_\ell - \partial_{x_j}u) f \mathrm{d}x $$
The second term on the RHS tends to zero using Cauchy-Schwarz and the assumed convergence of $v_\ell\to u$. Similarly we also have
$$ \lim_{\ell \to \infty}\int_U \partial_{x_j} w_\ell f \mathrm{d}x = \int_{U} \partial_{x_j} u f \mathrm{d}x $$
So we have that
$$ \int_U |Du|^2 \mathrm{d}x = \lim_{i,j\to \infty} \int_U Dv_i \cdot D w_j ~\mathrm{d}x \leq \lim_{i,j\to\infty} C \|v_i\|_{L^2} \|D^2 w_j\|_{L^2} $$
by (*). Since $v_i \to u$ in $H^1_0$, we also have $v_i \to u$ in $L^2$. Similarly as $w_j \to u$ in $H^2$ we have $D^2w \to D^2 u$ in $L^2$. So the RHS is
$$ \lim_{i,j\to\infty} C \|v_i\|_{L^2} \|D^2 w_j\|_{L^2} = C\|u\|_{L^2} \|D^2 u\|_{L^2}$$
and we have the desired result.
A partial answer.
In the book by Gilbarg and Trudinger it is proved the following (Theorem 8.12, page 186 of the last edition). I simplify the statement a little bit.
$f \in L^2(U)$ and $\varphi \in H^2(U)$ are given. If $L$ is strictly elliptic with smooth coefficients and the boundary of the domain $U$ is regular, then the solution of
$$
\begin{aligned}
&Lu = f \text{ in } U\\
&u-\varphi \in H^2(U)
\end{aligned}
$$
satisfies
$$
\|u\|_{H^2(U)}^2 \leq C(\|u\|_{L^2(U)}^2+\|f\|_{L^2(U)}^2+\|\varphi\|_{H^2(U)}^2).
$$
$C$ depends on $L,U$. The theorem is written without squared norms, but you can switch from one to the other easily.
Using this theorem it can be proved that (this is Evans exercise in the case $L = -\Delta$) if $u \in H^2(U) \cap H_0^1(U)$, then
$$
\|u\|_{H^2(U)}^2 \leq C(\|u\|_{L^2(U)}^2+\|-\Delta u\|_{L^2(U)}^2).
$$
Just set $f = \Delta u \in L^2(U)$. A solution of
$$
\begin{aligned}
&\Delta w = f \text{ in } U\\
&w = 0 \text { on } \partial U
\end{aligned}
$$
must satisfy
$$
\|w\|_{H^2(U)}^2 \leq C(\|w\|_{L^2(U)}^2+\|-\Delta w\|_{L^2(U)}^2).
$$
but $u$ itself is a solution of that problem. Actually the book by Evans is also a reference for this theorem.
To prove the exercise in general one should prove that
$$
\theta\|-\Delta w\|_{L^2(U)}^2 \leq (Lu,-\Delta u),
$$
where $\theta$ is $L$ ellipticity constant. This can be easily seen in dimension 1, but I did not do the computation in higher dimension. This last point should be checked.
Best Answer
That can be done, though it is a bit messy. In the following we let $f_{ij}$ denotes the partial derivative $\frac{\partial^2 f}{\partial x^i \partial x^j}$. Basically you want to show that for any $u$,
$$ \frac{d}{dt}\bigg|_{t=0} \mathcal F (u_t) = 0, $$
where $u_t = u+ t\varphi$ for any $\varphi \in C^\infty_0(\Omega)$ and $\mathcal F (f) = \int_{\Omega} (1+ |Df|^2)^{-a} \det (D^2 f) \ \mathrm dx$.
Direct calculations give
\begin{align} &\frac{d}{dt}\bigg|_{t=0} \mathcal F (u_t) \\ &=\int_\Omega \left(\frac{-2a \det (D^2 u)}{(1+ |du|^2)^{a+1}} \langle D u , D \varphi\rangle+ \frac{1}{(1+ |Du|^2)^a } ( \varphi_{11} u_{22} + \varphi_{22} u_{11} - 2\varphi_{12}u_{12})\right) \mathrm dx.\end{align}
For the second term, one use integration by part twice to get
$$\int_\Omega \left( \frac{\varphi_{11} u_{22} - \varphi_{12} u_{12} }{(1+|Du|^2)^a}\right) \mathrm dx = \int_\Omega \frac{2a\det D^2 u}{(1+ |Du|^2)^{a+1}} \varphi_1 u_1\mathrm d x.$$
and similarly
$$\int_\Omega \left( \frac{\varphi_{22} u_{11} - \varphi_{12} u_{12} }{(1+|Du|^2)^a}\right) \mathrm dx = \int_\Omega \frac{2a\det D^2 u}{(1+ |Du|^2)^{a+1}} \varphi_2 u_2\mathrm d x.$$
thus these two terms cancel the first term and thus the derivative is zero. This implies $\mathcal F(u)$ is invariant under any interior perturbation and thus depends only on its value on the boundary. Clearly it does not depend on $u$. But it is not clear to me why it depends only on $Du$ (even when $a=2$ or $3/2$).