Let $a \in L^2(\Omega)$ (bounded $\Omega$) with $0 \leq a(x) \leq C$ a.e.
We know $C^\infty(\overline{\Omega})$ is dense in $L^2(\Omega)$, so there exist smooth functions $a_n \to a$ in $L^2$.
But can we find a sequence $a_n$ such that $0 \leq a_n(x) \leq C$ (a.e)?
I think so. Because if $a_n \to a$ in $L^2$ for $a_n$ smooth, then for a subsequence, relabelled $a_n$, we have $a_n \to a$ pointwise a.e. This subsequence we can probably constrain to not exceed the bounds $(0,C)$...
John suggested one approach in a comment: convolution with a bump function preserves the inequalities such as $c\le a_n\le C$.
But maybe you are using another way to construct the initial sequence of smooth functions $a_k\to a$. In that case you can use smooth truncation by means of composition with a function $\phi : \mathbb R\to [c,C]$. For definiteness, let's consider $c=-1$, $C=1$. Consider the sequence of functions $$\phi_n(x) =\alpha_n^{-1} \int_0^x \frac{1}{1+t^{2n}} \, dt,\quad \text{where } \alpha_n = \int_0^\infty \frac{1}{1+t^{2n}} \, dt $$ which converges uniformly to $\max(1,\min(-1,x))$. Note that $|\phi_n|<1$.
Thus, for any smooth $a_k$, the composition $\phi_n\circ a_k$ is a smooth function, and $\phi_n\circ a_k \to \max(1,\min(-1,a_k))$ uniformly, hence in $L^p$.
Also, $$\| a - \max(1,\min(-1,a_k))\|_{L^p} \le \|a-a_k\|_{L^p} $$ because $|a|\le 1$; the integral on the left is smaller pointwise.
Thus, by choosing large $k$ and then large $n$, we get a smooth function approximating $a$ and bounded between $\pm 1$.