Let $X$ be a metric space consisting of a countable set of points, the distance between any two of which is $2$, together with one additional point $e$ whose distance to any of the other points is $1$. Then ${\rm Lip}_0(X)$ is isometrically isomorphic to $l^\infty$, which fails the RNP.
Another example: let $X = [0,1]$ with $e = 0$. Then ${\rm Lip}_0(X)$ is isometrically isomorphic to $L^\infty[0,1]$, which fails the RNP.
$\newcommand{\Om}{\Omega}\newcommand{\Th}{\Theta}\newcommand{\B}{\mathscr B}\newcommand{\M}{\mathcal M}\newcommand\ep\varepsilon\newcommand{\de}{\delta}\newcommand{\R}{\mathbb R}$By the polar decomposition of complex measures, there is a Borel function $g\colon\Om\to[0,2\pi)$ such that
\begin{equation*}
|\mu|(\Th)=\int_\Om 1_\Th e^{ig}\,d\mu. \tag{1}\label{1}
\end{equation*}
Take any real $\ep>0$. Next, take any natural
\begin{equation*}
m>\frac{2\pi |\mu|(\Th)}{\ep/2} \tag{2}\label{2}
\end{equation*}
and any
\begin{equation*}
\de\in\Big(0,\frac\ep{2(2m+1)}\Big). \tag{3}\label{3}
\end{equation*}
For $j\in[m]:=\{1,\dots,m\}$, let
\begin{equation*}
I_j:=[\tfrac{2\pi(j-1)}m,\tfrac{2\pi j}m),\ A_j:=\Th\cap g^{-1}(I_j), \tag{4}\label{4}
\end{equation*}
so that the $A_j$'s are Borel sets forming a partition of $\Th$.
Since $\Om$ is a metric space and $|\mu|$ is a Borel measure, $|\mu|$ is regular. So, for each $j\in[m]$ there exist a closed set $F_j$ and an open set $G_j$ such that
\begin{equation*}
F_j\subseteq A_j\subseteq G_j\text{ and }|\mu|(G_j\setminus F_j)<\de, \tag{5}\label{5}
\end{equation*}
so that the $F_j$'s are (pairwise) disjoint and for
\begin{equation*}
F:=\bigcup_{j\in[m]}F_j\text{ and }G:=\Th\setminus F \tag{6}\label{6}
\end{equation*}
we have
\begin{equation*}
|\mu|(G)=\sum_{j\in[m]}|\mu|(A_j\setminus F_j)<m\de. \tag{8}\label{8}
\end{equation*}
All metric spaces are normal. So, by Urysohn's lemma, for each $j\in[m]$ there exists a continuous function $h_j\colon\Om\to\R$ such that
\begin{equation*}
h_j=1\text{ on }F_j,\ h_j=0\text{ on }G_j^c:=\Om\setminus G_j,\ 0\le h_j\le1. \tag{9}\label{9}
\end{equation*}
Let
\begin{equation*}
h:=\sum_{j\in[m]} \frac{2\pi j}m\,h_j. \tag{10}\label{10}
\end{equation*}
Then, by \eqref{6}, \eqref{10}, \eqref{9}, and \eqref{4}, on $F$ we have $0\le h-g\le\frac{2\pi}m$, and hence
\begin{equation*}
|e^{ih}-e^{ig}|\le\frac{2\pi}m\quad \text{on}\ F. \tag{11}\label{11}
\end{equation*}
Again by the regularity of $|\mu|$ and Urysohn's lemma, there exist a closed set $F_0$ and a continuous function $h_0\colon\Om\to\R$ such that
\begin{equation*}
F_0\subseteq\Th,\ |\mu|(\Th\setminus F_0)<\de, \tag{12}\label{12}
\end{equation*}
\begin{equation*}
h_0=1\text{ on }F_0,\ h_0=0\text{ on }\Th^c,\ 0\le h_0\le1. \tag{13}\label{13}
\end{equation*}
So, by \eqref{1}, \eqref{13}, \eqref{6}, \eqref{11}, \eqref{12}, \eqref{8}, \eqref{2}, \eqref{3},
\begin{equation*}
\begin{aligned}
&\Big||\mu|(\Th)-\int_\Om h_0 e^{ih}\,d\mu\Big| \\
=&\Big|\int_\Th e^{ig}\,d\mu-\int_\Th h_0 e^{ih}\,d\mu\Big| \\
\le&\int_\Th |1-h_0|\,d|\mu|+\int_\Th|e^{ig}-e^{ih}|\,d|\mu| \\
=&\int_\Th |1-h_0|\,d|\mu|+\int_F|e^{ig}-e^{ih}|\,d|\mu| +\int_G|e^{ig}-e^{ih}|\,d|\mu| \\
\le&|\mu|(\Th-F_0)+ \frac{2\pi}m\,|\mu|(F)+2|\mu|(G) \\
\le&\de+ \frac{2\pi}m\,|\mu|(\Th)+2m\de<\ep.
\end{aligned}
\end{equation*}
So,
\begin{equation*}
\begin{aligned}
|\mu|(\Th)&=\Re|\mu|(\Th) \\
&\le\ep+\Re\int_\Om h_0 e^{ih}\,d\mu \\
&=\ep+\lim_n\Re\int_\Om h_0 e^{ih}\,d\mu_n \\
&=\ep+\liminf_n\Re\int_\Om h_0 e^{ih}\,d\mu_n \\
&=\ep+\liminf_n\Re\int_\Th h_0 e^{ih}\,d\mu_n \\
&\le\ep+\liminf_n|\mu_n|(\Th).
\end{aligned}
\end{equation*}
Letting $\ep\downarrow0$, we conclude that
\begin{equation*}
|\mu|(\Th)\le\liminf_n|\mu_n|(\Th),
\end{equation*}
as desired.
Best Answer
It is always true that $L_q(\mu,X^*)\hookrightarrow L_p(\mu,X)^*$ isometrically (without the assumption that $X$ has the Radon-Nikodym property). We need the Radon-Nikodym property to guarantee that this isometric embedding is a surjection.
From this it follows that for any $T\in L_q(\mu,X^*)$, $$\|T\|=\sup\Bigl\{\Bigl|\int \langle T,f\rangle d\mu\Bigr| : f\in B_{L_p(\mu,X)}\Bigr\}.$$ If $\int \langle T,f\rangle d\mu=0$ for all $f\in L_p(\mu,X)$, then it follows that $\|T\|=0$, and $T$ is the zero vector. From this it follows that $T=0$ $\mu$ a.e.