First, a simple case that $\Omega\subset\mathbb{R}$ and that $u(t,x)$ vanishes at a countable set of values $x_n(t)$, $n=1,2,\ldots$, with $\partial_x u(t,x)\neq 0$ at each $x_n(t)$.

Then
$$\partial_t\int_{\{u(t,\cdot) >0\} } 1\, dx=\partial_t\int_\Omega\theta[u(t,x)]\, dx$$
$$= ∫_\Omega \delta[u(t,x)]\partial_t u(t,x)\,dx = \sum_n\lim_{x\rightarrow x_n(t)}\frac{\partial_t u(t,x)}{|\partial_x u(t,x)|}.$$
In the first equality I introduced the unit step function $\theta(s)$, in the second equality I used that the derivative of the step function is a Dirac delta function, with the chain rule, in the third equality I used that $\delta[f(x)]=\sum_n \delta(x-x_n)/|f'(x_n)|$ if $f(x_n)=0$ with a nonvanishing first derivative $f'$. (See for example, Wikipedia.)

Now more generally, if $\Omega\subset\mathbb{R}^N$, let $S(t)$ be the $N-1$ dimensional surface on which $u(t,x)=0$, and assume that the gradient $\nabla_x u(t,x)$ does not vanish on $S(t)$. Then the desired result is a surface integral $\int_Sd\sigma(x)$,
$$\partial_t\int_{\{u(t,\cdot) >0\} } 1\, dx= ∫_\Omega \delta[u(t,x)]\partial_t u(t,x)\,dx=\int_{S(t)}\frac{\partial_t u(t,x)}{|\nabla_x u(t,x)|}d\sigma(x).$$
For the last identity, see for example Wikipedia.