$Hello$, Tomasz! (for some reason the MO prohibits saying "Hi" or "Hello" in the normal text mode). Nice to see you back. Apparently you are still asking the same question whether a function $H$ close to the product $fg$ can be represented as a product $FG$ where $F$ is close to $f$ and $G$ is close to $g$ but now just in the continuous category.
The answer is "Not necessarily" even for the closed unit disk $\mathbb D=\{|z|\le 1\}$. The obvious counterexample would be $f(z)=Mz,g(z)=M\bar z$, $H=M^2|z|^2+\varepsilon$ with huge $M$, but you tried to exclude it by demanding that $f$ and $g$ vanish nowhere. However, it doesn't save the day. Indeed, consider any non-negative continuous functions $\varphi, \psi:\mathbb D\to[0,1]$ such that $\varphi=0$ on $[0,1]$ and $\varphi=1$ outside a small neighborhood of $[0,1]$ while $\psi$ has the same properties with respect to the interval $[-1,0]$. Put $f(z)=Mz\varphi(z)-\frac{\varepsilon}{6M}, g(z)=M\bar z\psi(z)+\frac{\varepsilon}{6M}$. Then the product $fg$ is $\varepsilon$-close to a strictly positive function $H(z)=M^2|z|^2\varphi(z)\psi(z)+\frac\varepsilon 2$. However, if $H=FG$ and $F,G$ are $M/10$ close to $f$ and $g$ respectively, then the argument of $F$ should essentially follow that of $f$ on the left semicircle where $|f|$ is large and then that of $1/g$ on the right semicircle where $|g|$ is large (because $F=H/G$ and we control the arguments of both $H$ and $G$), i.e., it cannot deviate much from that of $z$ anywhere, so the winding number will be $1$ and $F$ will be forced to have a zero inside $\mathbb D$, which is impossible since $H>0$ everywhere.
Of course, if $X$ is the circle, this effect is excluded and the answer becomes "Yes", but you said that you knew it yourself, so I'll stop here for now.
