A Brownian motion is always defined with repect to a given probability space. Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and $X_t=W_t^\mathbb{P}$ a Brownian motion, i.e. a stochastic process with i.i.d. increments $X_t-X_s\sim N(0,t-s)$ and continuous sample paths $\mathbb{P}$-a.s. and with $X_0=0$.
Now, let $\mathbb{Q}\sim\mathbb{P}$ be a new probability measure defined on the measurable space $(\Omega,\mathcal{F})$. Due to the equivalence, the sample paths of $X_t$ are continuous $\mathbb{Q}$-almost surely but what about the distribution of the increments? $\mathbb{E}^\mathbb{P}[X_t-X_s]=0$ does not imply $\mathbb{E}^\mathbb{Q}[X_t-X_s]=0$. Thus, in general, $W_t^\mathbb{P}$ is not a Brownian motion anymore if you alter the probability measure and hence the associated expectation operator etc.
When you say that $W_t^\mathbb{Q}$ is a $\mathbb{Q}$-Brownian motion, you mean that it satisfies the definition with respect to the given probability space $(\Omega,\mathcal{F},\mathbb{Q})$. If you alter any component of the probability space, the process may not satisfy the original definition anymore.
Similarly, martingales are always defined with respect to a certain measure (expectation) and filtration. If you change the probability measure or the filtration, the considered process is not necessarily a martingale anymore.