Consider the Heaviside step function: $$H:\mathbb{R}\to \mathbb{R} $$ defined by $$H(x)=\begin{cases} 0 & \mbox{if } x<0 \\ 1 & \mbox{if } x\geq 0\end{cases}$$
Fix any $\delta>0$. Given any $\epsilon >0$, does there exist a real analytic function $F$ such that \begin{align} \sup_{\mathbb{R}}|F(x)-H(x)|<\epsilon\quad\quad \ldots (1) \end{align} and the complexification $\tilde{F}$ of $F$ has no singularities in $\{z\in\mathbb{C}:Im(z)<\delta\}$?
Note: One can use functions like $$F(x)=\frac{1}{1+e^{-kx}}$$ This should satisfy the proximity condition (1) for a sufficiently large $k$ but the singularity condition in the complex domain fails (Has a singularity at $z=\frac{\pi i}{k}$).
The answer is obviously no, even if you relax $F$ to be continuous.
Your requirement (1) asks for a sequence of continuous functions that approximate $H$ uniformly on $\mathbb{R}$. That is impossible since $H$ is not continuous.