$\def\R{\mathrel R}$No, this is not possible.
Recall that the depth of a point $x$ in a transitive frame $(W,R)$ is the maximal length $d$ of a strictly increasing chain starting at $x$, i.e., $x_1,\dots,x_d$ such that $x_d=x$ and $x_{i+1}\R x_i$, $x_i\not\R x_{i+1}$.
There are formulas in one variable that are satisfiable only in frames of depth $\ge d$ (cf. Thm. 12.21 in Chagrov&Zakharyaschev, Modal logic). Moreover, it is possible to define them in such a way that when satisfied in a model of depth exactly $d$, they force a particular value for $p$ in all points in the root cluster; we will obtain a contradiction from this. An explicit construction of such formulas follows below.
Consider the formulas
$$\begin{align}
\theta_1(p)&=\Box p,\\
\theta_{i+1}(p)&=p^{i+1}\land\Diamond\theta_i(p)\land\Box\Bigl(p^{i+1}\lor\bigvee_{j\le i}\theta_j(p)\Bigr),
\end{align}$$
where
$$p^i=\begin{cases}\phantom{\neg}p&\text{if $i$ is odd,}\\\neg p&\text{otherwise.}\end{cases}$$
We will use the property that these formulas are pairwise contradictory; moreover, the following formulas are valid:
$$\theta_j\to\Box\neg\theta_i,\qquad j<i.\tag{$*$}$$
We can prove this by induction on $i$. For $i=1$, there is nothing to prove. Assuming it holds for $i$, we show it for $i+1$ as follows. Let $j\le i$, and assume for contradiction that $x\R y$ are such that $x\models\theta_j$ and $y\models\theta_{i+1}$. If $j<i$, $y\models\Diamond\theta_i$ contradicts the induction hypothesis. If $j=i$, we have $y\models\neg p^i$. This directly contradicts the definiton of $\theta_1$ for $i=1$; otherwise, the definition of $\theta_i$ gives $y\models\bigvee_{j<i}\theta_j$, which together with $y\models\Diamond\theta_i$ contradicts the induction hypothesis again. This finishes the proof of $(*)$.
Now, assume for contradiction that $(W,R,w_0,V)$ is as in the question, and let $d$ be the depth of $w_0$. The formula $\theta_d(p)$ is satisfiable in $(W,R,w_0)$ by the valuation that makes $p$ true in points of odd depth, and false in points of even depth. That is, $\neg\theta_d(p)$ is not valid in the pointed frame $(W,R,w_0)$, thus by assumption, $w_0\not\models\Box\neg\theta_d(p)$, i.e., there is $x_d$ such that
$$w_0\R x_d\models\theta_d(p).$$
Unwinding the definition, we find a chain $x_d\R x_{d-1}\R\dots\R x_1$ such that $x_i\models\theta_i(p)$. This implies $x_i\not\R x_{i+1}$, as $\theta_i\to\Box\neg\theta_{i+1}$ is valid by $(*)$. Thus, the chain $x_d,\dots,x_1$ is strictly increasing. Since $w_0$ does not have depth $\ge d+1$, we must have $x_d\R w_0$. This implies $w_0\models p^d\lor\bigvee_{j\le d-1}\theta_j(p)$. Using $(*)$, we cannot have $w_0\models\bigvee_{j\le d-1}\theta_j(p)$ as $w_0\R x_d\models\theta_d(p)$, thus we obtain
$$w_0\models p^d.$$
However, since $\neg\theta_d(p)$ is not valid in the frame, $\neg\theta_d(\neg p)$ is not valid there either. Then the same argument as above with $p$ and $\neg p$ swapped gives
$$w_0\models\neg p^d.$$
This is a contradiction.
I formulated the argument above for reflexive transitive frames as requested, but it can be easily adapted to arbitrary finite pointed Kripke frames $(W,R,w_0)$: we take for $d$ the depth of $w_0$ under the transitive closure of $R$, and replace all instances of $\Box$ inside the $\theta_i$ formulas by the defined modality
$$\Box^{\le n}\phi=\bigwedge_{i=0}^n\underbrace{\Box\dots\Box}_i\phi,$$
where $n=|W|$. Note that $\Box^{\le n}$ is the box modality corresponding to the transitive reflexive closure of $R$.
