= Solution
Let $M$ bound the <operator norm> of $D^2F$; the bound on individual second derivatives supplies such an $M$ depending also on dimension. The <Taylor theorem> with integral remainder gives
$$
F(p)=F(0)+DF(0)\cdot p+\int_0^1(1-t)\,p^TD^2F(tp)p\,dt.
$$
Thus <quadratic growth from a bounded Hessian> gives
$$
|F(p)|\leq|F(0)|+|DF(0)||p|+\frac M2|p|^2\leq C(1+|p|^2).
$$
Since $\Omega$ is bounded it has finite measure, and a function in the <Sobolev space> $H^1(\Omega)$ has $Du\in L^2(\Omega)$. The given lower bound also implies that the integrand is nonnegative. Consequently
$$
\boxed{0\leq\mathcal F(u)=\int_\Omega F(Du)\leq C\left(|\Omega|+\|Du\|_2^2\right)<\infty.}
$$
For later use the same <Hessian> bound yields $|DF(p)|\leq|DF(0)|+M|p|$, hence $DF(Du)\in L^2(\Omega)$.
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