This is a well established concept in General Topology: «convergence structures». The two references I would recommend are the first chapters of each one of the following books:
E. Binz, Continuous Convergence on $C(X)$. LNM Springer, 469.
R. Beattie and H.-P. Butzmann, Convergence Structures and Applications to Functional Analysis.
A quick overview: On a set $X$, for every $x\in X$ we define which filters converge to $x$, with the following restrictions: the ultrafilter of all supersets of $x$ must converge to $x$; any filter which contains a filter converging to $x$ must converge to $x$; the intersection of two filters converging to $x$ must converge to $x$ («contains» and «intersection» to be understood in the usual set-theoretic sense).
Converging filters in a convergence subspace: a filter ${\mathcal F}$ converges to $x$ in the subspace iff the filter on the initial space generated by the filter basis ${\mathcal F}$ converges to $x$.
The sets which are present in every filter converging to $x$ are known as neighborhoods of $x$ with respect to the corresponding convergence structure $\Lambda$. Such sets are actually neighborhoods in the topological sense, for a topology on $X$ (called the topology associated with $\Lambda$). The definitions of a closed set as a set whose complementary is open, and as a set which coincides with its (filterwise) adherence, are equivalent. A set $K$ is said to be $\Lambda$-compact if every ultrafilter in $K$ converges with respect to the induced convergence structure on $K$. Filterwise continuous maps send compact sets onto compact sets.
$\newcommand{\bR}{\mathbb{R}}$ For any Polish space $S$ (separable complete metric space) we denote by $M_1(S)$ the space of Borel probability measures on $S$.
The space $\newcommand{\eP}{\mathscr{P}}$ $\eP:=M_1(\bR)$ is a topological space with respect to the weak convergence. In fact, $\eP$ with this topology is a Polish space.
A random measure $\mu$ on $\bR$ is by definition a measure on $\eP$, i.e.,
$$\mu \in M_1\bigl(\;\eP\;\bigr).$$
The space $C_b(\bR)$ embeds in $C_b\bigl(\eP\bigr)$ $\newcommand{\bsE}{\boldsymbol{E}}$
$$ f\in C_b(\bR)\mapsto \widehat{f}\in C_b\bigl(\eP\bigr), \;\;\widehat{f}(\pi):=\bsE_\pi(f)=\int_{\bR} f(t) \pi(dt),\;\;\forall \pi\in \eP. $$
wher $\bsE_\pi$ denotes the expectation of a random variable with respect to the probability distribution $\pi$. Denote by $\newcommand{\eX}{\mathscr{X}}$ $\eX\subset C_b\bigl(M_1(\bR)\bigr)$ the image of this embedding.
In your question you are given $\newcommand{\bmu}{\boldsymbol{\mu}}$ a sequence $\bmu_n\in M_1(\eP)$ and another measure $\bmu$ with the property that
$$\lim_n\int_{\eP}F(\pi) \bmu_n(d\pi)=\int_{\eP}F(\pi)\bmu(d\pi),\;\;\forall F\in \eX $$
and you ask if the above holds for all $F\in C_b(\eP)$. Clearly this happens iff $\eX$ is dense in $C_b(\eP)$. It is not hard to see that is not the case. Fix a nontrivial, compactly supported continuous function $\rho:\bR\to[0,1]$. Then the bounded, continuous function
$$ Q_\rho:\eP\to\bR,\;\;Q_\rho(\pi)=\left(\int_{\bR}\rho(t) \pi(dt)\,\right)^2, $$
is not in the closure of $\eX$.
Best Answer
To summarize the situation. Let $(X,\mathcal{F})$ a measurable space. The space $M(X,\mathcal{F})$ of all real-valued signed measures on $(X,\mathcal{F})$ is a Banach space wrto the total variation norm $\|\mu\|:=|\mu|(X)$ where the (non-negative) measure $|\mu|:= \mu_+ +\mu_-$ is the variation of $\mu$. The space $M(X,\mathcal{F})$may be isometrically embedded as a norm-closed subspace of the dual space $B(X,\mathcal{F})^*$ of the Banach space $B(X,\mathcal{F})$ of all bounded measurable functions on $X$, $B(X,\mathcal{F})$ with the uniform norm $\|\cdot\|_\infty$ ( that dual is in general much larger, since it also contains all additive measures). Note that the space of simple functions $S(X,\mathcal{F})$, linear span of characteristic functions of measurable sets, is norm dense in $B(X,\mathcal{F})$ via the usual approximation $f_n(x):= \lfloor nf(x)\rfloor/n$.
So the above "strong convergence of measures", that is with test functions in $B(X,\mathcal{F})$, is the weak* convergence of $B(X,\mathcal{F})^*$, in the particular cas eof sequences in $M(X,\mathcal{F})$. In particular any such convergent sequence of measures is norm bounded (as any w* convergent sequence of elements in a dual space), and it is "weakly convergent" in the above sense, that is with characteristic functions, hence also with simple functions as test. It is not norm convergent in general (as an example, take as said e.g. $\mu_n$ absolutely continuous w.r.to the Lebesgue measure on $X:=[0,1]$ and with densities $g_n\in L^1$ weakly convergent but not norm convergent).
Conversely, a norm bounded sequence of measures $\mu_n$ that weakly converges to $\mu$ wrto test functions $g$ in $S(X,\mathcal{F})$ also converges with test functions $f$ in $B(X,\mathcal{F})$, again an elementary and general fact. You can see it writing $$\langle \mu_n, f\rangle - \langle\mu, f\rangle=\langle\mu_n-\mu, g\rangle+ \langle\mu_n-\mu, g -f\rangle $$ with a simple function $g$, so that $$\limsup_{n\to\infty}|\langle\mu_n, f\rangle-\langle\mu, f\rangle|\le \limsup_{n\to\infty}|\langle\mu_n-\mu, g\rangle|+ \big(\sup_n \|\mu_n\|+\|\mu\|\big)\|g -f\|_\infty $$ whence $$\limsup_{n\to\infty}|\langle\mu_n, f\rangle-\langle\mu, f\rangle|=0,$$
since simple functions are $\|\cdot\|_\infty$ dense.