5.9.
$$\begin{array}{l} \text { Let } x_{1}=-\log _{2014} \frac{b}{a}, x_{2}=-\log _{2014} \frac{c}{b}, \\ x_{3}=-\log _{2014} \frac{d}{c} . \end{array}$$

Since $a>b>c>d>0$, we have
$$x_{1}>0, x_{2}>0, x_{3}>0 \text {. }$$

Thus, the given inequality can be transformed into
$$\begin{array}{l} \frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}} \leqslant m \cdot \frac{1}{x_{1}+x_{2}+x_{3}} \\ \Rightarrow m \geqslant\left(x_{1}+x_{2}+x_{3}\right)\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}\right) \geqslant 9 . \end{array}$$

When $x_{1}=x_{2}=x_{3}$, i.e., $a, b, c, d$ form a geometric sequence, the equality holds.
Therefore, the minimum value of $m$ is 9.