Certainly not. For $m=4$, $\lfloor 4 x_4 \rfloor = 23$. It is true that $m=9$ works.
Numerical experiments (using floating-point approximations as the exact values of $x_n$
rapidly get out of hand) seem to show that $x_{m^2}$ grows more slowly than $m$ as $m$ increases. I suspect that suitable bounds could prove this.
EDIT: Consider the iteration $x_{n+1} = x_n + \dfrac{a}{x_n}$ where $x_1 = 1$ and $0 < a \le 1$.
Note that $$x_{n+1}^2 = x_n^2 + 2 a + \frac{a^2}{x_n^2} \le x_n^2 + 2 a + a^2$$
By induction, $x_n \le \sqrt{1 + (2a+a^2) (n-1)}$.
In our case we have $a = \dfrac{1}{\sqrt m - 1}$, so
$$ m x_{m^2} \le m\sqrt {1+ \left( \left( \sqrt {m}-1 \right) ^{-2}+2\, \left( \sqrt {
m}-1 \right) ^{-1} \right) \left( {m}^{2}-1 \right) }
$$
and this is less than $m^2$ if $m > (2 + \sqrt{2})^2 \approx 11.65$.
Conclusion: After eliminating other values $\le 11$, $m=1$ and $m=9$ are the only solutions.
Proof for $p ≥ 1$
Since $u^p - 1 ≥ p(u - 1)$ for all $u ≥ 0$, it suffices to prove the result for $p = 1$. That follows from
$$\frac{x^y}{y} - 1 ≥ \frac{1 + y \ln x}{y} - 1
= \ln x + \frac1y - 1 ≥ \ln x + \ln \frac1y = \ln x - \ln y$$
by cyclic summation over $(x, y) = (x_i, x_{i + 1})$.
Conjectured proof for $p ≥ \frac12$
Since $u^p - 1 ≥ 2p(u^{\frac12} - 1)$ for all $u ≥ 0$, it suffices to prove the result for $p = \frac12$. Numerical evidence suggests that
$$\left(\frac{x^y}{y}\right)^{\frac12} - 1 ≥ \frac{\ln x}{2\sqrt[4]{1 + \frac13 \ln^2 x}} - \frac{\ln y}{2\sqrt[4]{1 + \frac13 \ln^2 y}}$$
for all $x, y > 0$. If this is true, cyclic summation yields the desired result.
Counterexample for $0 < p < \frac12$
Let $g(x) = \left(\frac{x^{1/x}}{1/x}\right)^p + \left(\frac{(1/x)^x}{x}\right)^p$. Then $g(1) = 2$, $g'(1) = 0$, and $g''(1) = 4p(2p - 1) < 0$, so we have $g(x) < 2$ for $x$ in some neighborhood of $1$. This yields counterexamples for all even $n$:
$$\left(x, \frac1x, x, \frac1x, \dotsc, x, \frac1x\right), \quad x ≈ 1, \quad 0 < p < \frac12.$$
For $n = 3$, the best counterexample seems to be
$$(0.41398215, 0.73186577, 4.77292996), \quad 0 < p < 0.39158477.$$
Best Answer
Use $\mod n$ for subscripts and call the expression you want to maximize $X$.
If $2\mid n$, by simple transformation, \begin{align*}X={}&\sum_{k=1}^n\left(\frac{a_k}2-a_ka_{k+1}+\frac{a_{k+1}}2\right)\\[2pt]={}&\sum_{k=1}^n\left(-\frac{\left(2a_k-1\right)\left(2a_{k+1}-1\right)-1}4\right).\end{align*} Since $2a_k-1$ and $2a_{k+1}-1\in[-1,1]$, each term of the sum $\le\dfrac12$. So $X\le\dfrac n2=\left\lfloor\dfrac n2\right\rfloor$.
If $2\nmid n$, notice that $X$ is a linear function of $a_n$, so maximum is reached at $a_n=0$ or $1$. If $a_n=0$, also $a_1a_n=0$, then \begin{align*}X={}&\sum_{k=1}^{(n-1)/2}\left(a_{2k-1}+a_{2k}-a_{2k-1}a_{2k}-a_{2k}a_{2k+1}\right)\\[2pt]\le{}&\sum_{k=1}^{(n-1)/2}\left(a_{2k-1}+a_{2k}-a_{2k-1}a_{2k}\right)\\[2pt]={}&\sum_{k=1}^{(n-1)/2}\left(1-(1-a_{2k})(1-a_{2k-1})\right)\\[2pt]\le{}&1\cdot\frac{n-1}2=\frac{n-1}2=\left\lfloor\frac n2\right\rfloor.\end{align*} If $a_n=1$, then $a_1a_n$ cancels with $a_1$. So \begin{align*}X={}&\sum_{k=1}^{(n-1)/2}\left(a_{2k}+a_{2k+1}-a_{2k-1}a_{2k}-a_{2k}a_{2k+1}\right)\\[2pt]\le{}&\frac{n-1}2=\left\lfloor\frac n2\right\rfloor.\end{align*} This inequality can be proved by a nearly identical method with the one demonstrated in the last case.