For a ring $R$, the space $\operatorname{MaxSpec} R$ -- the maximal ideals of $R$ endowed with the Zariski topology -- has two evident properties: its singleton sets are closed (i.e., the **T1** separation property) and it is quasi-compact.  In [his thesis][1], M. Hochster showed that conversely any quasi-compact T1 topological space $X$ is homeomorphic to $\operatorname{MaxSpec} R$ for some commutative ring $R$.  

A totally disconnected topological space is automatically T1: closures of points are connected sets!  So your question reduces to one of general topology: is there a totally disconnected, quasi-compact non-Hausdorff topological space $X$?  The answer is yes and indeed [Eric Wofsey's answer to your earlier math.stackexchange.com question][2] provides an example.

In summary: **no, not necessarily**.




[1]: http://www.ams.org/journals/tran/1969-142-00/S0002-9947-1969-0251026-X/S0002-9947-1969-0251026-X.pdf
[2]: http://math.stackexchange.com/questions/1507986/is-totally-disconnected-space-hausdorff/1508024#1508024