Supremum of difference of Brownian bridges: strictly positive wp 1? 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.
 A: The answer is $1$, even if $B_1$ and $B_2$ are not independent. Indeed, take any $u>0$. Then 
\begin{align*}
 P( \sup_{x\in\mathbb{R}} [B_1(F_1(x)) - B_2(F_2(x))] \ge -u )
 &\ge \sup_{x\in\mathbb{R}}P(B_1(F_1(x)) - B_2(F_2(x)) \ge -u ) \\ 
 &\ge \lim_{x\to-\infty}P(B_1(F_1(x)) - B_2(F_2(x)) \ge -u ) \\ 
 &=P(B_1(0) - B_2(0) \ge -u )=P(0-0 \ge -u )=1,
\end{align*}
because $B_i(\cdot)$ is continuous and therefore $B_i(t)\underset{t\downarrow0}\longrightarrow B_i(0)=0$ almost surely and hence in probability and in distribution. 
So, 
\begin{equation}
 P( \sup_{x\in\mathbb{R}} [B_1(F_1(x)) - B_2(F_2(x))] \ge 0 )=
 \lim_{u\downarrow0}P( \sup_{x\in\mathbb{R}} [B_1(F_1(x)) - B_2(F_2(x))] \ge -u )
 =\lim_{u\downarrow0}1=1. 
\end{equation}

The OP has changed the question, now asking if $P( \sup_{x\in\mathbb{R}} [B_1(F_1(x)) - B_2(F_2(x))] > 0 )=1$. In general, the answer to this question is no: consider e.g. the case when $B_1=B_2$ and $F_1=F_2$. 
However, the answer remains yes if, as is assumed in the OP's question, $B_1$ and $B_2$ are independent. The main idea here is that the Brownian bridge is close to the corresponding Brownian motion on any small interval of the form $[0,t]$. 
Indeed, we can write 
$B_i(t)=W_i(t)-tW_i(1)$, where $W,W_1,W_2$ are independent standard Brownian motions. 
Let 
\begin{gather*}
 X(x):=B_1(F_1(x)) - B_2(F_2(x)),\quad Y(x):=W_1(F_1(x)) - W_2(F_2(x)),\\
 Z(x):=Y(x)-X(x)=F_1(x)W_1(1)-F_2(x)W_2(1). 
\end{gather*}
Note that the process $(Y(x))_{x\in\mathbb{R}}$ equals the process  $(\sqrt2\,W(\frac{F_1(x)+F_2(x)}2))_{x\in\mathbb{R}}$ in distribution and 
\begin{equation}
 \lim_{x\to-\infty}\frac{Z(x)}{\sqrt{F_1(x)+F_2(x)}}=0. 
\end{equation}
So, the probability in question is 
\begin{align*}
 P( \sup_{x\in\mathbb{R}} [B_1(F_1(x)) - B_2(F_2(x))] >0 )
 &\ge P( \limsup_{x\to-\infty} \frac{X(x)}{\sqrt{F_1(x)+F_2(x)}} =\infty) \\
  &=P( \limsup_{x\to-\infty} \frac{Y(x)-Z(x)}{\sqrt{F_1(x)+F_2(x)}} =\infty) \\ 
  &=P( \limsup_{x\to-\infty} \frac{Y(x)}{\sqrt{F_1(x)+F_2(x)}} =\infty) \\  
  &=P( \limsup_{x\to-\infty} \sqrt2\,\frac{W(\frac{F_1(x)+F_2(x)}2)}{\sqrt{F_1(x)+F_2(x)}} =\infty)=1 
\end{align*}
by the law of the iterated logarithm -- see e.g. Corollary 5.3 in M\"orters and Peres. 
So, $P( \sup_{x\in\mathbb{R}} [B_1(F_1(x)) - B_2(F_2(x))] >0 )=1$. 
A: (Edited: I added additional details)
The answer is yes also with the strict inequality. Denote by $a_j = \inf \{x : F_j(x) > 0\} \in [-\infty, \infty)$ the left endpoint of the support of $F_j$, $j = 1, 2$. With no loss of generality assume that $a_1 \leqslant a_2$.
With probability one, there is a decreasing sequence $(t_n)$ which converges to $0$ and such that $B_1(t_n) > 0$. Note that $t_n$ are random, but they depend only on $B_1$. Write $x_n = F_1^{-1}(t_n)$ and $s_n = F_2(x_n)$. Then $(x_n)$ converges to $a_1$. Furthermore, if $a_1 = a_2$, then $s_n > 0$ and $(s_n)$ converges to $0$; if $a_1 < a_2$, then $s_n = 0$ for $n$ large enough.
Conditioning on $B_1$, we obtain
$$ \operatorname{Pr}(\sup_{x \in \mathbb{R}}[B_1(F_1(x)) - B_2(F_2(x))] > 0) \geqslant \operatorname{Pr}(\exists n : B_2(s_n) \leqslant 0) . $$
The right-hand side is one, because (as we prove below) for any decreasing sequence $(s_n)$ which converges to zero we have $\operatorname{Pr}(\forall n : B_2(s_n) > 0) = 0$. Note that we can consider the sequence $(s_n)$ to be fixed, because it only depends on $B_1$.
If $a_1 < a_2$, then $s_n = 0$ for $n$ large enough, and so obviously $\operatorname{Pr}(\forall n : B_2(s_n) > 0) = 0$. Hence, we suppose that $a_1 = a_2$, so that $s_n > 0$ and $(s_n)$ converges to $0$.
We choose a (random) subsequence $s_{k_n}$ recursively in such a way that
$$ \operatorname{Pr}(B_2(s_{k_{n+1}}) \geqslant 0 | B_2(s_{k_1}), \ldots, B_2(s_{k_n})) < \tfrac{3}{4} \tag{1}$$
(see below for a remark why this is possible). Then we have
$$ \operatorname{Pr}(B_2(s_{k_1}) \geqslant 0, \ldots, B_2(s_{k_n}) \geqslant 0) < (\tfrac{3}{4})^n , $$
and so
$$ \operatorname{Pr}(\forall n : B_2(s_{k_n}) \geqslant 0) = 0 , $$
as desired.
To see why it is possible to choose $s_{k_{n+1}}$ so that (1) is satisfied, note that for $m > k_n$, the conditional distribution of $B_2(s_m)$ given the values of $B_2(s_{k_1}), \ldots, B_2(s_{k_n})$ is normal with mean $(s_m / s_{k_n}) B_2(s_{k_n})$ and variance $s_m (s_{k_n} - s_m)$. It follows that the conditional probability in (1) is equal to
$$
  \Phi\left(\frac{(s_m / s_{k_n}) B_2(s_{k_n})}{\sqrt{s_m (s_{k_n} - s_m)}} \right) ,
$$
where $\Phi$ is the CDF of the standard normal distribution. As $m \to \infty$, the above expression converges to $\Phi(0) = \tfrac{1}{2}$, so it is less than $\tfrac{3}{4}$ when $m$ is large enough.
