EDIT: the original $\ge$ is now $>$ (sorry for the typo!)

Let $B_1(\cdot)$ and $B_2(\cdot)$ denote independent, standard Brownian bridges, i.e., they are mean-zero Gaussian processes on $[0,1]$ with $B(0)=B(1)=0$ and covariance function $t_1(1-t_2)$ for $t_1 \le t_2$. Let $F_1(\cdot)$ and $F_2(\cdot)$ each be a cumulative distribution function (CDF) that is continuous. (CDFs are also non-decreasing and satisfy $\lim_{x\to-\infty}F(x)=0$ and $\lim_{x\to\infty}F(x)=1$.)

What is $\Pr( \sup_{x\in\mathbb{R}} [B_1(F_1(x)) - B_2(F_2(x))] > 0 )$?

My guess/special case: I think the answer should be one. For example, if $F_1(\cdot)=F_2(\cdot)=F(\cdot)$, then $B_1(F_1(\cdot))-B_2(F_2(\cdot)) = B_1(F(\cdot)) - B_2(F(\cdot))$, so \begin{align} \Pr( \sup_{x\in\mathbb{R}} [B_1(F_1(x)) - B_2(F_2(x))] > 0 ) &=\Pr( \sup_{x\in\mathbb{R}} [B_1(F(x)) - B_2(F(x))] > 0 ) \\ &=\Pr( \sup_{t\in[0,1]} [B_1(t) - B_2(t)] > 0 ) \\ &=\Pr( \sup_{t\in[0,1]} \sqrt{2} B(t) > 0 ) \\ &= 1 . \end{align} The second-to-last line follows since $B_1(\cdot)$ and $B_2(\cdot)$ are independent, their difference has the distribution $\sqrt{2} B(\cdot)$ for some other standard Brownian bridge $B(\cdot)$. The final line follows from, e.g., Theorem 2 in Smirnov (1939) or equation (1.1) in Birnbaum and Tingey (1951).

However, if $F_1 \ne F_2$, then I don't think the difference $B_1(F_1(\cdot))-B_2(F_2(\cdot))$ is technically a Brownian bridge (it is some other sort of "Gaussian bridge"?), so I do not know how to rigorously show that the probability is one (if indeed it is!), nor do I have a reference for such a result.

Also: the answer below (to the original version with my typo...) shows that the probability of being $\ge0$ is one, so the question is now equivalent to the question whether the probability that the supremum equals zero is zero.