How long would it take for sucrose to undergo hydrolysis in boiling water?
This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer.
Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions.  In general this is so that the spontaneous rate can be compared to the enzyme-catalyzed rate, so that the catalytic proficiency of enzymes can be calculated.
All that's just prefatory remarks to introduce this paper, &quot;Rates of Spontaneous Cleavage of Glucose, Fructose, Sucrose, and Trehalose in Water, and the Catalytic Proficiencies of Invertase and Trehalase&quot;.
The abstract's first sentence is:
The half-lives for spontaneous hydrolysis of trehalose and sucrose at 25 °C are 6.6 × 106 years and 440 years.
Thus, a solution of sucrose in water should be 50% decomposed into monomers after 440 years -- at 25 °C.  That value comes from thermodynamic extrapolation of experiments done at higher temperatures.  (I don't think they wanted to wait hundreds of years to do an experiment at 25 °C.)
But higher temperatures is exactly what you're after.  Figure 1 from the paper has the data.

The lower-right datapoint is very close to 100 °C, and shows the first-order rate constant for sucrose hydrolysis is about $10^{-6}$ per second.  That means that 63% degradation would take $10^{6}$ seconds, or about 11.5 days.  Getting 99% degradation would take about 53 days.
edits
The above has been edited to fix sloppy math errors, as noted by @gremin and @John Bentin in the comments.  Thank you both.