The model companion $U$ of a theory $T$ in a language $L$ is a model completion if for every model $M$ of $T$, the union of $U$ and the diagram of $M$ is a complete theory in the language $L(M)$. It is well known that a model companion is a model completion iff $T_\forall$, the set of universal consequences of $T$, has the amalgamation property (AP).
Can one go further and say that a theory $T$ has a model completion only if $T_\forall$ has the strong amalgamation property (SAP)? (A theory has the SAP if the AP for $A_1 \hookleftarrow A_0 \hookrightarrow A_2$ may be witnessed in such a way that the intersection of the images of $A_1$ and $A_2$ is the image of $A_0$.) Famous examples of theories having the AP but not the SAP, such as those of distributive lattices and fields of positive characteristics, do not exactly fit into this picture, I don't think.
Let $T$ be the theory of algebraically closed fields. Then $T_\forall$ is the theory of integral domains. $T$ is the model completion of itself (and also of $T_\forall$), but $T_\forall$ does not have SAP.
For an even more elementary example, let $T$ be the universal theory of an equivalence relation $E$ such that every $E$-class has at most $2$ elements. Then $T$ does not have SAP (let $A_0 = \{a\}$, $A_1 = \{a,b\}$ with $a\neq b$ and $aEb$, and $A_2 = \{a,c\}$ with $a\neq c$ and $aEc$), but $T$ has a model completion: the theory of an equivalence relation $E$ with infinitely many classes, all of size $2$.