The answer is yes, because every forcing notion is equivalent to a forcing notion with finite predecessors.
Theorem. Every forcing notion is forcing equivalent to a forcing notion with the finite-predecessor condition.
Proof. Consider any forcing notion $\newcommand\P{\mathbb{P}}\P$. Let $\P^*$ be the forcing notion consisting of the finite pointed subsets of $\P$, that is, finite sets $a\subset\P$ such that $a$ has a least element. We order $\P^*$ by $a\leq b$ if and only if $b\subseteq a$. Since a finite set has only finitely many subsets, this will ensure that $\P^*$ has the finite-predecessor condition.
It is easy to see that $\P^*$ projects to $\P$ by mapping every pointed set to its point. So forcing with $\P^*$ adds a generic for $\P$.
Conversely, I claim that forcing with $\P$ adds a generic for $\P^*$. Assume that $G\subset\P$ is $V$-generic, and let $G^*$ consist of the finite pointed subsets of $G$. This is a filter in $\P^*$, since $G$ is a filter in $\P$. This is $V$-generic for $\P^*$, since if $D^*\subset\P^*$ is dense, then let $D$ consist of the least points of any pointed set in $D^*$ — this will be dense in $\P$. So for any $a\in G^*$ with least point $p\in G$, there is some $q\in D$ which is least in some $b\in D^*$ such that $q\in G$. So $b\in G^*$ and so $G^*$ meets $D^*$.
Since $G$ and $G^*$ are easily constructed from each other, we have $V[G]=V[G^*]$, and so these forcing notions are forcing equivalent. $\Box$
It seems that every forcing notion is thus equivalent to a forcing notion with finite conditions.