[For convenience, I will pretend in this answer that any compactification actually contains $X$ as a subspace, so I don't have to constantly be writing down the embedding maps.]
No. For instance, you can define a compactification $Y=X\cup\{\infty\}$ where the only neighborhood of $\infty$ is the entire space (and every open subset of $X$ remains open). There will not exist any morphism from this compactification to $X^*$ unless the topology on $X^*$ happens to be the same as the topology on $Y$ (i.e., the only compact closed subset of $X$ is the empty set; given your assumption that $X$ is KC and noncompact, this is impossible!).
A separate issue is that $X$ is open in $X^*$, so if you have some other compactification in which $X$ is not open, you cannot expect it to have a morphism to $X^*$. There are also uniqueness issues--a morphism to $X^*$ does not need to send all the new points to $\infty$ (for instance, if $X$ is uncountable with the cocountable topology, you could let $X'$ be $X$ together with one more point with the cocountable topology and let $Y$ be the 1-point compactification of $X'$, and then the new point of $X'$ can map to anywhere in $X^*$ and the map will still be continuous). With non-Hausdorff spaces, a continuous map is not determined by its values on a dense subset, so generally it will be very hard to get any sort of uniqueness property like this without stronger hypotheses.
If you restrict your definition of "compactification" to require $Y$ to also be KC and that $X$ is open in $Y$, then it is true that $X^*$ is the terminal compactification (assuming $X^*$ is a compactification at all by this definition--it won't always be KC). These hypotheses make it trivial to check that the map $Y\to X^*$ sending every new point to $\infty$ is continuous (the hypothesis that $X$ is open in $Y$ gives continuity at points of $X$, and the hypothesis that $Y$ is KC gives continuity at new points).
For uniqueness, suppose $h:Y\to X^*$ is a morphism of compactifications, and let $A=X\cup h^{-1}(\{\infty\})\subseteq Y$. Then I claim $A$ is compact. To prove this, note that $h^{-1}(\{\infty\})$ is closed in $Y$ and hence compact, so it suffices to show any ultrafilter $F$ on $X$ has a limit in $A$. By compactness of $Y$, $F$ has a limit $y\in Y$; if $y\in A$ we're done, so we may assume $y\not\in A$. In that case $h(y)\neq\infty$, so it is a point of $X$, and then since $h$ is the identity on $X$, $F$ must converge to $h(y)$ in $X$. Thus $h(y)$ is a limit of $F$ in $A$.
Thus since $Y$ is KC, $A$ is closed in $Y$. Since $A$ contains $X$ and $X$ is dense in $Y$, this means $A=Y$. Thus $h$ must map every point of $Y\setminus X$ to $\infty$.
For a direct topological proof, suppose $f:X\to Y$ is not surjective and let $y\in Y\setminus f(X)$. In a Stone space, points can be separated by clopen sets, and then by a compactness argument points and closed sets can be separated by clopen sets. So, there is a clopen set $C\subseteq Y$ such that $y\in C$ and $f(X)\subseteq Y\setminus C$. We now have $1_Cf=0f$ where $1_C,0:Y\to\{0,1\}$ are the characteristic function of $C$ and the constant $0$ map, respectively. This witnesses that $f$ is not epic in the category of Stone spaces.
Best Answer
Yes, according to Herrlich & Strecker, Section 6.10(4). Here’s the argument:
That’s as far as they actually write it out, but clearly the rest is that $h\circ f=k\circ f$, and $f$ is an epimorphism, so $h=k$, and $f[A]$ must therefore be dense in $B$.
Added: To see that $C$ is actually Hausdorff, let the copies of $B$ be $B_0=B\times\{0\}$ and $B_1=B\times\{1\}$, let $K=\operatorname{cl}f[A]$, and let $K_i=K\times\{i\}$ for $i\in\{0,1\}$. Finally, let $q:B_0\sqcup B_1\to C$ be the quotient map. Clearly $q(\langle x,i\rangle)$ and $q(\langle y,j\rangle)$ can be separated by disjoint open sets in $C$ whenever $x\ne y$, irrespective of $i$ and $j$. If $q(\langle x,0\rangle)\ne q(\langle x,1\rangle)$, then $x\in B\setminus K$, an open subset of $B$, so $q[B_0\setminus K_0]$ and $q[B_1\setminus K_1]$ are disjoint open nbhds of $q(\langle x,0\rangle)$ and $q(\langle x,1\rangle)$.