For any given value of $p$, consider the probability of a queen being able to traverse the board vertically as being $q\left(p\right)$, and the probability of a rook being able to traverse the board horizontally (ie. from left edge to right edge) as being $r\left(p\right)$.  This notation is to emphasise that these values are dependent on $p$.

The key observation, as I see it, is that for any given board, exactly one of the conditions "queen can traverse the board vertically" and "rook can traverse the board horizontally using the missing squares of the board" is true. If the rook has a horizontal path across the missing squares, then this path cuts off the queen's access to a vertical path.  This "virtual board" consisting of the missing squares has fraction $1-p$ of its squares intact.

Next, notice that every board with fraction $p$ of the squares intact has a "transpose board" (flipped along the leading diagonal) which has the exact same fraction $p$ intact.  Any piece which can traverse a board vertically can traverse its transpose board vertically, and vice-versa.  This means for a given $p$, the fraction of boards the rook can traverse horizontally is the same as the fraction of boards it can traverse vertically.

Thus (again for fixed $p$) the probability $r(1-p)$ of a rook being able to traverse a random board's complement horizontally is the same as the probability of a rook being able to traverse a random board's complement vertically. But we know that exactly one of "a queen can traverse the board vertically" and "a rook can traverse the board horizontally on the missing squares" is true, so $r(1-p)+q\left(p\right)=1$. 

This gives us $q\left(p\right)=1-r(1-p)$.  In the neighbourhood of the critical value $p_{queen}$, we have $q(p_{queen}+ϵ)\approx 1$, and $q(p_{queen}-ϵ)\approx 0$, so $r\left(\right(1-p_{queen})-ϵ)\approx 0$ and $r\left(\right(1-p_{queen})+ϵ)\approx 1$. But this is only true if $(1-p_{queen})$ is the critical value for a rook, ie. $p_{queen}+p_{rook}=1$.

I suppose the reason you asked for both probabilities together is that it helps to consider when either or both pieces can or cannot traverse certain boards, as I did.  It would have been significantly more effort, if not impossible to deduce the probability of either individually without consideration as to the other.

We use the same argument as with the queen and the rook, but instead consider the bonsdman and antibondsman. By symmetry, they have the same probabilities, and the bondsman can traverse a board vertically if and only if an antibondsman cannot traverse the board's complement horizontally.  This means for the same reason as before, $2\cdot p_{bondsman}=p_{bondsman}+p_{antibondsman}=1$, ie. $p_{bondsman}=\frac{1}{2}$.

With regard to your bonus question, I have a hunch that this is somehow related to the bond problem in percolation theory (somehow, your title may have provided a hint here).  Maybe this is the transition / critical probability for when bonds form in some specific type of lattice.

Considering squares as representing nodes here and the bondsman's ability to travel between two squares to be an edge, this means every square is linked to exactly six squares (four cardinal directions and the two possible diagonals).  This gives six edges, which indicates that we might be dealing with a cubic lattice?