Here is the long version of my comment.

First, let us show that if $X$ is any $1D$ strong Markov process with $X_0 = 0$ which is symmetric under sign inversion (i.e the Markov semigroup commutes with the composition operator with $x \mapsto -x$) and has continuous trajectories, then its variance cannot decrease. The proof very closely follows Yuri's argument. Let $\delta>0$ be fixed and construct a process $Y_t$ with the same law as $X_{t+\delta}$ as follows. $Y$ runs independently from $X$ until the stopping time $$ \tau = \inf\{t > 0\,:\, |Y_t| \ge |X_{t+\delta}|\}\;. $$ For $t > \tau$, we then set $Y_t = X_{t+\delta}$. Since trajectories are continuous, one has $|Y_\tau| = |X_{\tau + \delta}|$ and $|Y_t| \le |X_{t+\delta}|$ for every $t$. By the strong Markov property, $Y$ has the same law as $X$ and so we have shown that $|X_{t}|$ is stochastically dominated by $|X_{t+\delta}|$ for every $t, \delta > 0$, which is more than we need.

Another counterexample: The above construction relied on the strong Markov property to conclude that $Y$ and $X$ are equal in law. If one drops this, the conclusion fails in general even in the $1D$ symmetric case. Consider the following (slightly crazy) example. Let $\mu$ be the random measure on the dyadics in $[0,1]$ which is such that $\mu(\{k/2^n\})$ is exponentially distributed with mean $3^{-n}$ whenever $k$ contains no factor $2$. Furthermore, these exponential random variables are all independent. Let also $Z$ be the solution to the SDE $$ dZ = -10 Z\,dt + Z\,dW\;,\qquad Z_0 = 1\;. $$ (The only properties we need are that the law of $Z$ has a nice density w.r.t. Lebesgue, only charges positive numbers, has continuous trajectories, and converges to $0$ in mean square as $t \to \infty$.) Finally, take a random coin toss $\sigma \in \{\pm1\}$ independent of all of the above. We then set $X_t = \sigma \bar X_t$, where $\bar X_t$ is built as follows. Write $\tau = \mu([0,1])$ for the total mass of $\mu$. For $t \le \tau$, we then set $$ \bar X_t = \inf\{y\ge 0\,:\, \mu((0,y]) \ge t\}\;. $$ Loosely speaking, $\bar X$ runs through all dyadics in increasing order and, when it is located at $k 2^{-n}$ it jumps to "the next" dyadic with rate $3^{-n}$. For $t \ge \tau$, we then set $\bar X_t = Z_{t-\tau}$. Obviously, the variance of $X$ is not monotone. It is a nice exercise to show that $X$ has continuous trajectories (easy) and is Markov but not strong Markov (harder). Hint: Dyadics have measure $0$ under the law of $Z_t$ for any fixed $t$.

Here is the long version of my comment.

First, let us show that if $X$ is any $1D$ strong Markov process with $X_0 = 0$ which is symmetric under sign inversion (i.e the Markov semigroup commutes with the composition operator with $x \mapsto -x$) and has continuous trajectories, then its variance cannot decrease. The proof very closely follows Yuri's argument. Let $\delta>0$ be fixed and construct a process $Y$ with the same law as $X$ as follows. $Y$ runs independently from $X$ until the stopping time $$ \tau = \inf\{t > 0\,:\, |Y_t| \ge |X_{t+\delta}|\}\;. $$ For $t > \tau$, we then set $Y_t = X_{t+\delta}$. Since trajectories are continuous, one has $|Y_\tau| = |X_{\tau + \delta}|$ and $|Y_t| \le |X_{t+\delta}|$ for every $t$. By the strong Markov property, $Y$ has the same law as $X$ and so we have shown that $|X_{t}|$ is stochastically dominated by $|X_{t+\delta}|$ for every $t, \delta > 0$, which is more than we need.

Another counterexample: The above construction relied on the strong Markov property to conclude that $Y$ and $X$ are equal in law. If one drops this, the conclusion fails in general even in the $1D$ symmetric case. Consider the following (slightly crazy) example. Let $\mu$ be the random measure on the dyadics in $[0,1]$ which is such that $\mu(\{k/2^n\})$ is exponentially distributed with mean $3^{-n}$ whenever $k$ contains no factor $2$. Furthermore, these exponential random variables are all independent. Let also $Z$ be the solution to the SDE $$ dZ = -10 Z\,dt + Z\,dW\;,\qquad Z_0 = 1\;. $$ (The only properties we need are that the law of $Z$ has a nice density w.r.t. Lebesgue, only charges positive numbers, has continuous trajectories, and converges to $0$ in mean square as $t \to \infty$.) Finally, take a random coin toss $\sigma \in \{\pm1\}$ independent of all of the above. We then set $X_t = \sigma \bar X_t$, where $\bar X_t$ is built as follows. Write $\tau = \mu([0,1])$ for the total mass of $\mu$. For $t \le \tau$, we then set $$ \bar X_t = \inf\{y\ge 0\,:\, \mu((0,y]) \ge t\}\;. $$ Loosely speaking, $\bar X$ runs through all dyadics in increasing order and, when it is located at $k 2^{-n}$ it jumps to "the next" dyadic with rate $3^{-n}$. For $t \ge \tau$, we then set $\bar X_t = Z_{t-\tau}$. Obviously, the variance of $X$ is not monotone. It is a nice exercise to show that $X$ has continuous trajectories (easy) and is Markov but not strong Markov (harder). Hint: Dyadics have measure $0$ under the law of $Z_t$ for any fixed $t$.