I assume "strongly measurable" is in the sense of Bochner. I define nonnegative measurable functions $f_\epsilon$ for my example. See below$^*$ for a modification with positive measurable functions.
Let $f_\epsilon$ be defined by $$ f_\epsilon(x) = \begin{cases} 1,\quad&\text{ if }0<x<\epsilon<1 \\ 0,\quad&\text{ otherwise} \end{cases} $$ so $f^*(x) = \sup_{\epsilon > 0}f_\epsilon(x) = \mathbf1_{(0,1)}(x)$ and therefore $f^* \in L_p(\mathbb R)$.
Define $F : \mathbb R \to \ell_{(0,\infty)}^\infty$ by
$F(x) = (f_\epsilon(x))_{\epsilon \in (0,\infty)}$.
Note: for $0<x<y<1$ we have
$$
\|F(y) - F(x)\|_\infty
= \sup_{\epsilon\in(0,\infty)} |f_\epsilon(y) - f_\epsilon(x)|
\ge |f_y(y) - f_y(x)| = 1 .
$$
So the range of $F$ (even if we omit a set of $x$ with measure zero) is nonseparable. Thus $F$ is not strongly measurable.
$^*$The above example has nonnegative functions $f_\epsilon$. For a similar example with positive measurable functions, we may do this: choose a fixed positive measurable function $g \in L^p$ and consider $g+f_\epsilon$. Then for $G(x) = ((g+f_\epsilon)(x))_{\epsilon \in (0,\infty)}$. We have $$ \|G(y) - G(x)\|_\infty = \|F(y) - F(x)\|_\infty $$ so we get the same conclusion that $G$ is not strongly measurable.