OK, here is the extended version of my comment.

Let $u(x,y)$ be the (bounded) harmonic function in $D = (-1,1) \times (0, \infty)$, with boundary value $1$ along $\{-1,1\} \times (0, \infty)$ and $0$ along $(-1,1) \times \{0\}$. Consider a 2-D Brownian motion $(X_t, Y_t)$, started at $(X_0,Y_0) = (x,y)$, and denote the corresponding probability law by $P^{x,y}$. Either by Itô's lemma, or by potential-theoretic results, $u(X_t,Y_t)$ is a martingale up to $\tau_D$, the first time $(X_t,Y_t)$ hits the boundary of $D$. In particular,
$$u(x,y) = E^{x,y} u(X_{\tau_D},Y_{\tau_D}).$$
At $t = \tau_D$, we either have $X_t = \pm 1$ or $Y_t = 0$. Let $\sigma$ be the first time $Y_t = 0$, and $\tau$ be the first time $X_t = \pm 1$. Clearly, $\tau_D = \min\{\sigma, \tau\}$. If $\sigma < \tau$, then $\tau_D = \sigma$ and $Y_{\tau_D} = 0$. Thus, $u(X_{\tau_D},Y_{\tau_D}) = 0$. On the other hand, if $\sigma > \tau$, then $X_{\tau_D} = \pm 1$, and hence $u(X_{\tau_D},Y_{\tau_D}) = 1$. It follows that
$$u(x,y) = P^{x,y}(\sigma > \tau).$$
By symmetry and translation invariance, with the notation of the problem,
$$u(0,y) = P(\sigma_y > \tau_1).$$
By self-similarity (a.k.a. scaling),
$$u(0,\sqrt{r}) = P(\sigma_{\sqrt{r}} > \tau_1) = P(\sigma_r > \tau_{\sqrt{r}}).$$

By the boundary Harnack inequality, we know that a finite, positive limit
$$ \lim_{y \to 0^+} \frac{u(0, y)}{y} $$
exists. This implies that
$$P(\sigma_r > \tau_{\sqrt{r}}) \sim c \sqrt{r}$$
as $r \to 0^+$.

As a side remark: In fact, there is no need to employ BHI here, as soon as one observes that the formula $u(x,-y) = -u(x,y)$ extends $u$ to a harmonic function in $(-1, 1) \times \mathbb{R}$. It is then clear that $u(0,y)$ is a smooth function of $y$, and it remains to show that $\partial_y u(0, y) > 0$. This last stem follows for example from the explicit expression for the Poisson kernel of a strip; however, I think one can easily cook up a soft argument here.

By the way, Hopf's lemma provides a yet another proof of the asymptotic formula for $u(0, y)$.

optimal, your estimate indeed seemsoptimistic, or even super-optimistic: unless I got something wrong, by scaling, we have $P(\sigma_r > \tau_{r^{1/2}}) = P(\sigma_{r^{1/2}} > \tau_1) \to 1$ as $r \to 0^+$. $\endgroup$ – Mateusz Kwaśnicki Oct 29 '19 at 9:08