Looking at the last paragraph of your question, it might seem reasonable to reformulate the question as: can we always select weights $\mathbf{w}$, such that the values $Q(V_i(\mathbf{w}))$ satisfy prescribed ratios: if we normalize by the sum of all entries $$s(\mathbf{w}):=\sum_{i=1}^nQ(V_i(\mathbf{w}))$$ we get a vector $$Q_{\text{norm}}(\mathbf{w}):=\frac{1}{s(\mathbf{w})}(Q(V_i(\mathbf{w})))_{1\leq i\leq n}$$ which is contained in the simplex $\Delta_{n-1}=\{(x_i)_{1\leq i\leq n}|\sum_i=1\}$. Then the question is: can we find for every point $p\in\Delta_{n-1}$ weights $\mathbf{w}$ such that $Q_{\text{norm}}=p$. For example if we take $p$ to be the barycenter of $\Delta_{n-1}$, then we would want to select weights such that $Q(V_i(\mathbf{w}))$ are equal for all $i$.
I can answer this question completely under the additional assumption that $Q$ is continuous. (Your examples "area, perimeter, diameter, or width of the cells" are all continous). What we need to prove is the surjectivity of the continous map $$Q_{\text{norm}}\colon\, \mathbb{R}^n\to \Delta_{n-1}.$$ We consider a map $$\begin{align}f\colon\, \Delta_{n-1}&\to\mathbb{R}^n\\x=(x_1,\dots,x_n)&\mapsto f(x)=(\log(x_1),\dots,\log(x_n)).\end{align}$$ This is only defined on the interior of $\Delta_{n-1}$, but we extend the definition on the boundary $\partial\Delta_{n-1}$ by setting $\log(0)=-\infty$. Also we extend the definition of the Voronoi diagram to allow some (but not all) components of the weight vector to be $-\infty$; the coresponding Voronoi regions will be empty. These definitions then allow us to compose $f$ with $Q_{\text{norm}}$ to get the map $$Q_{\text{norm}}\circ f\colon\, \Delta_{n-1}\to\Delta_{n-1},$$ which is continous and has the nice properties to map faces to faces: If some coordinates of $x$ are zero, the corresponding coordinates in $f(x)$ are $-\infty$ and the corresponding coordinates in $Q_{\text{norm}}\circ f(x)$ are also zero, since the corresponding Voronoi regions are empty. Therefore a face of $\Delta_{n−1}$ is mapped to itself.
Then we can apply a little (but often useful) lemma from algebraic topology that just states that such a map must be surjective. See for this question (an the answers): Map from simplex to itself that preserves sub-simplices. Hence $Q_{\text{norm}}\circ f$ is surjective and therefore also $Q_{\text{norm}}$ is surjective, which is what we wanted to show.
Edit: If $Q$ is taken to be the area and you are looking for an all equal area partition, then this partition is even unique up to sets of measure zero. A nice proof of this geometrical proof is given in the following paper, where they also consider some generalizations of the question:
(It is not the first proof of this results, see the references of this paper.)
