These measures do converge weakly to a measure with two atoms (of equal probabilities $1/2$) at the two paths $\phi_{\pm}:t\mapsto \pm t$. One way to see it is to consider $\frac{1}{r}B_t$, so that the condition is $||\cdot||_\infty\geq 1$, and apply [Schilder's theorem][1]: observe that $\phi_{\pm}$ are the only minimizers of the rate function $I(\phi)=\int_0^1 \phi'(t)^2\,dt$ in the set $S_0:=\{\phi:||\phi||_\infty\geq 1,\,\phi(0)=0\}$. On the other hand, for every positive $\epsilon>0$, we have $I((1+\epsilon)\phi_+)=(1+\epsilon)^2$, and an open $\epsilon$-neighborhood of $(1+\epsilon)\phi_+$ is contained in $S_0$, hence  
$$\liminf\frac{1}{r^2}\log\mathbb{P}\left(\frac{1}{r}B\in S_0\right)\geq -1.$$

Since this rate function is good, in particular, lower semi-continuous under $||\cdot||_\infty$ norm, we have that the sequence of conditioned measures is tight: for every $\epsilon>0$, the set $E_{1+\epsilon}:=\{\phi:I(\phi)\leq 1+\epsilon\}$ is compact, and 
$$
\limsup\frac{1}{r^2}\log\mathbb{P}\left(\frac{1}{r}B\in E^c_{1+\epsilon}|\frac{1}{r}B\in S_0\right)\leq \limsup\frac{1}{r^2}\log\mathbb{P}\left(\frac{1}{r}B\in E^c_{1+\epsilon}\right)-\liminf\frac{1}{r^2}\mathbb{P}\left(\frac{1}{r}B\in S_0\right)\leq -\inf_{\phi\in \overline{E_{1+\epsilon}}}I(\phi)+1= -\epsilon,
$$
in particular, $\mathbb{P}\left(\frac{1}{r}B\in E^c_{1+\epsilon}|\frac{1}{r}B\in S_0\right)\to 0$, proving tightness (at the end, we used lower continuity again).

On the other hand, for any closed set $F\subset S_0\setminus\{\phi_{\pm}\}$, we have, by lower semi-continuity again, $F\subset E^c_{1+\epsilon}$ for $\epsilon$  small enough. Therefore, portmanteau theorem implies that any subsequential limit gives probability $0$ to any such $F$, which by regularity of measures is enough to conclude that any subsequential limit is supported on $\phi_{\pm}$.


 


  [1]: https://en.wikipedia.org/wiki/Schilder%27s_theorem