For a ring $R$, the space $\operatorname{MaxSpec} R$ -- the set of 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**.

<b>Added</b>: On the other hand, $\operatorname{Spec} R$ -- the set of prime ideals of $R$ endowed with the Zariski topology -- is T1 iff it is Hausdorff iff it is totally disconnected iff $\operatorname{Spec} R = \operatorname{MaxSpec} R$.  See e.g. $\S$13.3 [of these notes][3].  Maybe this is helpful for you.


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