Dynamic voting quorum Problem:
Suppose that a committee with $n$ members needs to vote on whether to accept a proposition. 
Each member in the committee can cast a yes/no vote ($q_i\in\{1,0\}$ for $i\in \{1,2,...,n \}$), and each member's vote has a  different weight ($w_i\in[0,1]$ for $i\in \{1,2,...,n \}$).
The committee rules stipulate that the proposition is to be accepted if the weighted average of votes is more than 50%, namely, if $$\bar{q}(n)=\sum_{i=1}^{n} w_i q_i>0.5$$
Suppose that (i) voting is sequential, (ii) each vote is a random draw from a Bernoulli distribution with unknown mean $p$, and (iii) the order of voting is independent of voting weights. 
I would like to define a 'dynamic quorum rule', such that a decision (accept vs reject) is made with only a subset of the committee's votes (e.g. the first $k$ votes),  and yet be statistically confident that the decision made is the same as the one that would have been made if all members had voted.
What I currently have:
This problem is very similar to one that I posted earlier. The only difference is that I am now interested in a weighted solution. That means that for $w_i=w$, the bayesian solution proposed in the related question also applies here.
 A: Exact treatment (intractable?)
You're asking for an answer similar to the one accepted in your related question, but that approach seems unlikely to be tractable here; however, the development is sketched below in case someone else might have an insight.
Let $$S_{j,k}=\sum_{i=j}^k w_i q_i$$ where $1\le j\le k\le n$ and $q_1,...,q_n$ are i.i.d. Bernoulli($p$) random variables. The sequential decision procedure is as before, mutatis mutandis, using the following posterior probabilities:
$$\begin{align}p_k &= P\{S_{1,n}<acc\mid (q_i)_{i=1..k}\}\\
&= \sum_{t\ <\ acc}P\{S_{1,n}\!=\!t\mid S_{1,k}\!=\!s\}. 
\end{align}$$
Using a Beta$(\alpha,\beta)$ prior for $p$, as before, now leads to the following form for these posterior probabilities:
$$\begin{align}&P\{S_{1,n}=t\mid S_{1,k}=s\}\\
&=\int_0^1 P\{S_{1,n}\!=\!t\mid S_{1,k}\!=\!s, p\!=\!p'\}\,\frac{P\{S_{1,k}\!=\!s\mid p\!=\!p'\}f(p')}{\int_0^1 P\{S_{1,k}\!=\!s\mid p\!=\!p''\}f(p'')dp''}\,dp'\\ \\
&=\frac{\int_0^1 P\{S_{k+1,n}=t-s\mid p\!=\!p'\}\,P\{S_{1,k}\!=\!s\mid p\!=\!p'\}f(p')\,dp'}{\int_0^1 P\{S_{1,k}\!=\!s\mid p\!=\!p''\}f(p'')\, dp''}\\ \\
&= \frac{\int_0^1\ \ \sum_{A\subseteq\{w_{k+1},...,w_n\}\\ \sigma A=t-s}p^{\lambda A}(1-p)^{n-k-\lambda A}\ \ \sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}p^{\lambda A'}(1-p)^{k-\lambda A'}\ \ f(p)\,dp}{\int_0^1\ \ \sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}p^{\lambda A'}(1-p)^{k-\lambda A'}\ \ f(p)\,dp}\\ \\
&= \frac{\sum_{A\subseteq\{w_{k+1},...,w_n\}\\ \sigma A=t-s}\sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}\int_0^1 p^{\alpha+\lambda A+\lambda A'}(1-p)^{\beta+n-\lambda A-\lambda A'}\,dp}{\sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}\int_0^1 p^{\alpha+\lambda A'}(1-p)^{\beta+k-\lambda A'}\,dp}\\ \\
&= \frac{\sum_{A\subseteq\{w_{k+1},...,w_n\}\\ \sigma A=t-s}\sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}\mathrm{B}(\alpha+\lambda A+\lambda A',\,\beta+n-\lambda A-\lambda A')}{\sum_{A'\subseteq\{w_1,...,w_k\}\\ \sigma A'=s}\mathrm{B}(\alpha+\lambda A',\beta+k-\lambda A')}\tag{*}
\end{align}$$
where, as before, $\mathrm{B}(\alpha,\beta)$ denotes the standard complete Beta function. In the above, we have used the combinatorial fact that, for $1\le j\le k\le n$,
$$P\{S_{j,k}=s\mid p\} = \sum_{A\subseteq\{w_j,...,w_k\}\\ \sigma A=s}p^{\lambda A}(1-p)^{k-j+1-\lambda A} = \sum_{A\in \mathcal{P}(\{w_j,...,w_k\})\\ \sigma A=s}p^{\lambda A}(1-p)^{k-j+1-\lambda A}
$$
where $\sigma A$ denotes the sum of the elements in $A$, $\lambda A$ denotes the number of elements in $A$, and $\mathcal{P}(\{w_j,...,w_k\})$ denotes the power set of $\{w_j,...,w_k\}$. 
It can be seen that in the case of equal weights (e.g. $w_i=1$), $(*)$ reduces to the previous result -- the outer summation in the numerator provides the factor $\binom{n-k}{t-s}$ and each of the other two summations (the inner one in the numerator and one in the denominator) provide a factor of $\binom{k}{s}$ that cancel one another, while in the summations $\lambda A = t-s$ and $\lambda A'=s$. However, in the case of arbitrary weights the combinatorics make it difficult to proceed any further.
