What is the probability that two independent random vectors with a given euclidean distance $r$ fall in the same orthant? Consider two independent and identically distributed random vectors of dimensionality $N$, $\mathbf{x}$ and $\mathbf{y}$, where their elements are iid generated from a Gaussian with zero mean and variance equal to $\sigma^2$; i.e., $\mathbf{x} , \mathbf{y} \sim \mathcal{N}(0, \sigma^2)$. 
What is the probability that $\mathrm{sign}(\mathbf{x})=(\mathrm{sign(x_1)},\mathrm{sign(x_2)}, \ldots, \mathrm{sign(x_N)})$ be equal to  $\mathrm{sign}(\mathbf{y})=(\mathrm{sign(y_1)},\mathrm{sign(y_2)}, \ldots, \mathrm{sign(y_N)})$ if their Euclidean distance is smaller than $r$, $r \geq 0$: $\Pr\left[ \mathrm{sign}(\mathbf{x})=\mathrm{sign}(\mathbf{y})\ \vert\ \Vert \mathbf{x} - \mathbf{y} \Vert \leq r\right]$.
Note: $\mathbf{x}$ and $\mathbf{y}$ are independent.
 A: This is not a complete answer, but may provide you a way to go. As a hands-on-approach I suggest to perform a simulation to get an idea how the resulting property looks like and then start to derive a formula for it. 
Here is such a simulation for a fixed standard deviation. Playing around with other standard deviations returned similar results.
require(lattice)

stdev <- 1
N <- 5
trials <- 2000

resPerN <- data.frame("N"=rep(1:N,each=trials),
              "r"=rep(-1,N*trials),
              "prob"=rep(-1,N*trials))


for(n in 1:N){
    res <- data.frame("r"=rep(-1,trials),"signum"=rep(-1,trials))

    for(i in 1:trials){
      x <- rnorm(n,0,stdev)
      y <- rnorm(n,0,stdev)
      diff <- sqrt(sum((x-y)^2))
      signum <- sum(sign(x) == sign(y)) == n
      res[i,] <- c(diff,signum)
    }

    res <- res[order(res$r),]

    range <- (1+(n-1)*trials) : (n*trials)
    resPerN$N[range] <- n
    resPerN$r[range] <-  (res$r - min(res$r))/(max(res$r)-min(res$r))

    invprob <-  1/(cumsum(res$signum)/cumsum(1:trials))
    invprob[which(invprob == Inf)] <- 0
    resPerN$prob[range] <- (invprob - min(invprob))/(max(invprob)-min(invprob))
}

xyplot(prob~r,data=resPerN,groups=resPerN$N,type="b",xlab="(min-max-transformed) r",ylab="(min-max-transformed) 1/prob | <= r",auto.key=T,
par.settings=list(superpose.line = list(col = rainbow(5),lty = 1),
          superpose.symbol=list(col = rainbow(5),pch=15,cex=0.8)))

which resulted in this plot

which indicates a (n inverse) logistic relationship. 
Going further by collecting beta-coefficients and intercepts, one pair per dimension, you should be able to derive a function dependent on r. Afterwards I'd vary the standard deviation to include it into the equation.
