If $A$ is compact then $\frac fg$ will be uniformly continuous because as long as $g \
ot= 0$ on $A$, $\frac 1g$ will be continuous, and $\frac fg$ is then the product of two continuous functions on a compact set, hence uniformly continuous.
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$A$ merely closed isn't enough. For instance, $f(x) = x$ and $g(x) = \frac{1}{x}$ are both uniformly continuous on $[1,\infty)$, but $\frac fg(x) = x^2$ is not.