Consider these reactions:
\begin{equation}
\ce{HA + H2O <=> A- + H3O+}\tag{1}
\end{equation}
\begin{equation}
\ce{A- + H2O <=> HA + OH-}\tag{2}
\end{equation}
The equilibrium constant for reaction (1) is called "acid dissociation constant", $K_{\rm a}$, and the second is "base association constant", $K_{\rm b}$. (Note that the names are there for historical reasons, they're not correct, strictly speaking) We can factor out the water's concentration to a good approximation in dilute solutions. Consider that
\begin{equation}K_{\rm a} \times K_{\rm b} = \frac{\ce{[A- ][H3O+ ]}}{\ce{[HA]}} \times \frac{\ce{[HA][OH- ]}}{\ce{[A- ]}} = \ce{[H3O+ ][OH- ]} = K_{\rm w}\tag{3}\end{equation}
$K_{\rm w}$ is constant in constant temperature, hence $K_{\rm a}$ and $K_{\rm b}$ are inversely proportional to each other. Again, to a good approximation, we have
\begin{equation}-\log{K_{\rm a}} + (-\log{K_\rm b})=-\log{K_\rm w}\: \Longrightarrow {\rm p}K_{\rm a} + {\rm p}K_{\rm b} = 14\tag{4}\end{equation}
p$K_{\rm a}$s between -1.7 ($\ce{H3O+}$/$\ce{H2O}$) and 15.7 ($\ce{H2O}$/$\ce{OH-}$) are said to be those of weak acids in water. Acids with $K_{\rm a}$s in this range do not dissociate fully in water, and named weak acids (and weak bases, for that matter1) in water.
Now, think of it like this: $x+y=14$. If $x$ is 5, $y$ is 9, and if $x$ is -5, $y$ is 19. (4) is where your rule of thumb originates from. So
- A weak acid like acetic acid with a p$K_{\rm a}$ of 4.76 will have a weak conjugate base with a p$K_{\rm b}$ of $14-4.76=9.24$.
- A strong acid like $\ce{HCl}$ with a p$K_{\rm a}$ of -7 will have a conjugate base with a p$K_{\rm b}$ of $14-(-7)=21$.
Thus as Oscar's answer points out, it's inaccurate the way your guideline is phrased, but it's naturally following if we use comparatives for phrasing. I'd dare say most conjugate bases of weak acids mentioned in textbooks are weak ones.
1: As the comment points out, the weak base range is p$K_{\rm b}$>1. That slipped.