As for the first question the answer is no even if $\mu$ is assumed to be $\sigma$-finite as by a simple change of measure you may reduce the problem to the case where $\mu$ is finite. Let us fix a finite measure $\mu$.
When $p=1$ we may use the following reasoning. The space $L_1(\mu)$ is weakly compactly generated because the inclusion map ${\rm id}\colon L_2(\mu)\to L_1(\mu)$ has dense range. On the other hand, it is well-known that $\ell_1(\Gamma)$ does not embed into a weakly compactly generated space for any uncountable set $\Gamma$, however by the principle of local reflexivity you will find it in some ultrapower of $L_1[0,1]$.
When $p\in (1,2)$, then $\ell_1(\Gamma)$$\ell_p(\Gamma)$ does not embed into $L_p(\mu)$ with $\mu$ finite for $\Gamma$ uncountable; this is proved in:
P. Enflo, H. P. Rosenthal, Some results concerning $L_p(\mu)$-spaces. J. Functional Analysis 14 (1973), 325–348.
This was extended to $p\in (2,\infty)$ by Bill Johnson and Gideon Schechtman (see this paper).
As for the second question, should such construction exist, it is most likely covered in one of Fremlin's Measure theory volumes.