Queens. Two queens suffice, as pointed out in the comments of Sp3000.
Rooks. Three rooks suffice, by trapping the black king between two walls, and then using the third rook to slowly close the gap between the walls. If the black king approach a rook, simply move it on the same file but further away. Two rooks do not suffice, since there is no check-mate position with two white rooks and a black king.
Bishops. Six bishops suffice, with three of each color. White can use a pair of bishops to form a wall, and thus with two pairs white can trap the black king between two walls, which gradually close and deliver checkmate with the fifth or sixth (whichever color is needed). If the black king approaches a bishop, simply move it away on the same diagonal. Five bishops do not suffice, since the black king can simply stay on the color having at most two bishops of that color, and there will always be a square available, since double check will not arise.
Knights. No finite number of white knights will suffice in this game. To see this, observe first that two white knights do not suffice, since one cannot set up a check-mated position. Now, suppose white places finitely many knights on the board. Black need only place his king on a file completely to the right of those pieces. Black's strategy is simply to move steadily to the right (that is, his every move should involve a rightward horizontal movement). Since knights move at most two units to the right, white will not be able to keep more than two knights in the vicinity of the black king, and indeed, will not be able to bring two knights so close. With one knight, to be sure, white can move at double speed and catch up to the black king; but to bring two knights close to the king, each would have had to move at a greater than one unit to the right speed, which is not possible. Finally, notice that with infinitely many knights, we can fill up the board almost completely, leaving only a few empty spots, and easily set a trap for black this way.
Conclusion. So we have $Q=2$, $R=3$, $B=6$ and $N=\infty$, giving $$\frac{1}{Q}+\frac1R+\frac1B+\frac1N=\frac12+\frac13+\frac16+\frac1{\infty}=1.$$
For a $6 \times 3$ board, it's impossible. Consider the board below, and the squares, marked with an $\times$.
$$\begin{matrix}
\cdot & \cdot & * & \cdot & \cdot & \cdot \\
\times & \cdot & \cdot & \cdot & \times & \cdot \\
\cdot & \cdot & * & \cdot & \cdot & \cdot \\ \end{matrix}$$
The only way to get to these squares is through the squares marked with an $\ast$. But if you use both to get to one of them and to get out again, then you cannot reach the other marked square anymore. The only way to solve this is to start at one of these marked squares. By symmetry, we have two other such squares:
$$\begin{matrix}
\cdot & \cdot & \cdot & \ast & \cdot & \cdot \\
\cdot & \times & \cdot & \cdot & \cdot & \times \\
\cdot & \cdot & \cdot & \ast & \cdot & \cdot \\ \end{matrix}$$
So we have to finish at another of these squares as well. So the first four and final four moves are determined, as, one of the following two scenarios:
$$\begin{matrix}
\cdot & \cdot & 2 & 15 & \cdot & \cdot \\
1 & 16 & \cdot & \cdot & 3 & 18 \\
\cdot & \cdot & 4 & 17 & \cdot & \cdot \\ \end{matrix} \qquad \text{or} \qquad \begin{matrix}
\cdot & \cdot & 2 & 17 & \cdot & \cdot \\
1 & 16 & \cdot & \cdot & 3 & 18 \\
\cdot & \cdot & 4 & 15 & \cdot & \cdot \\ \end{matrix}$$
In both cases, we need the remaining $10$ squares to connect the two ends of the chain to form a knight's path. This means that each of these $10$ points is connected to two other of these points (except for the ones next to $2$ and $17$). By starting with the corners, we get two chains of length $5$:
$$\begin{matrix}
d & a & \cdot & \cdot & A & D \\
\cdot & \cdot & c & C & \cdot & \cdot \\
b & e & \cdot & \cdot & E & B \end{matrix}$$
There's just no way to connect $a/e$ to $A/E$, so there's no chain of length $10$ connecting these squares. So there is no knight's path on this board.
Best Answer
This is just illustration for what @Henning Makholm described:
(if somebody is interested in the implementation of this simulation, the code is here)
It looks that closed form of number of squares on infinite chessboard reachable at <= n knight's moves from a fixed square is known as A018836, and is following:
$$K_n = 1-6*n+14*n^2+4*sign(n*(n-1)*(n-3))$$.