I assume "strongly measurable" is [in the sense of Bochner][1].  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.



  [1]: https://en.wikipedia.org/wiki/Bochner_measurable_function