I know that if we have Brownian motion ($W$) on canonical space (denote the probability measure by $P$) and let $Q$ be the unique probability measure such that $$Q(A) = E_P\left(\mathbf{1}_A \exp \left(\alpha W_t - \frac{\alpha^2 t}{2} \right) \right) \quad \forall A \in \mathcal{F}_t$$ then $$\widetilde W_t \equiv W_t - \alpha t$$ is Brownian motion under $Q$. In order to use Girsanov's to get this conclusion, we needed that $Q$ and $P$ are equivalent (i.e. $P(A) = 0 \iff Q(A) = 0$). However, by the SLLN for Brownian motion, $$ \lim_{t \rightarrow \infty} \frac{W_t}{t} = 0 \quad P\text{ - a.s.} \quad \quad \text{and} \quad \quad\lim_{t \rightarrow \infty} \frac{\widetilde W_t}{t} = 0 \quad Q\text{ - a.s.} \implies \lim_{t \rightarrow \infty} \frac{W_t}{t} = \alpha \quad Q\text{ - a.s.}$$
My question is: how does this not contradict the equivalence of the probability measures $P$ and $Q$?
Girsanov's theorem says that if we let $Z_t := \exp(\alpha W_t - \frac 12 \alpha^2 t)$ and define $Q$ by $dQ = Z_T dP$, then $\tilde W_t := W_t - \alpha t$ is a Brownian motion on $[0,T]$ under $Q$. Note the statement of Girsanov's theorem typically includes the limitation that $\tilde W_t$ is only a Brownian motion on some fixed interval. Now, for each $T > 0$ we could define a measure $Q_T$ that is equivalent to $P$ by $dQ_T = Z_T dP$, and under each of these measures we would have that $\tilde W_t := W_t - \alpha t$ is a Brownian motion on $[0,T]$. Furthermore, the family $(Q_T)_{T > 0}$ is a consistent family of probability measures and we could extend them to a probability measure $\hat Q$ under which $\tilde W_t$ is a Brownian motion on $[0,\infty)$, but this measure would no longer be equivalent to $P$ for exactly the reason you stated.
In terms of what you wrote, it's true that $P(A)=0$ iff $Q(A)=0$ for all $A \in \mathcal F_t$, but the event $\{\lim_{t \rightarrow \infty} \frac{W_t}{t} = 0\}$ is not in $\mathcal F_t$ for any $t$ so this does not contradict the equivalence of the $P$ and $Q$ measures.