For any set $X$, we may consider the partial order $\mathbb{P}$ consisting of all finite partial functions from $\omega$ to $X$, ordered by extension. If $G\subset\mathbb{P}$ is $V$-generic for this forcing notion, then $f=\cup G$ is a function from $\omega$ onto $X$. It is a function, since conditions in $G$ are compatible; it is total, since it is dense to add any given point to the domain; and it is surjective, since for any $x\in X$ the set of conditions $p$ with $x\in\mathop{ran}(p)$ is dense. Thus, in the forcing extension $V[G]$, the set $X$ becomes countable. This is how any set can be made countable by forcing. Since the forcing notion $\mathbb{P}$ is a set, it also follows that $V[G]\models ZFC$.
In the case that $X$ is an arbitrary proper class, one can still form the partial order $\mathbb{P}$ consisting of all finite partial functions from $\omega$ to $X$, and it will still be true that the union of the generic filter will be a function from $\omega$ to $X$, thereby making $X$ countable in $V[G]$. But what will no longer be true in this class forcing case is that $V[G]\models ZFC$. Indeed, if $X$ is a proper class, then the corresponding forcing extension will definitely not satisfy ZFC, since $X$ will contain members of arbitrarily large ordinal rank (appearing unboundedly high in the $V_\alpha$ hierarchy), the forcing extension will have a function from $\omega$ unbounded in the ordinals, in violation of the Replacement axiom. So the price you pay for making a proper class $X$ countable is that you must give up ZFC in the forcing extension. Indeed, if $X$ is a proper class in $V$, then $X$ contains elements of unboundedly high rank, and this will remain true in any extension of $V$ to a model $W$ with the same ordinals; thus, $X$ will not be a set in any such $W$.
So, to answer your question: every set can be countable in a forcing extension by set forcing (and thus while preserving ZFC), but no proper class can become a set in any extension of the universe to a model of ZF with the same ordinals.
Edit. I see now that you want to consider upward extensions of the models, not just forcing extensions. There are a few observations to make about this. I have been interested in this topic for some time.
The question is the extent to which a model $M$ of set theory can become the $(V_\theta)^N$, for some taller of a taller model of set theory.
First, I claim that every model $M$ of set theory can be elementarily embedded into the $V_\theta$ of another model of set theory $N$, and you can even arrange that $(V_\theta)^N\prec N$. This is a sense in which $M$ is continued as you desire to a taller model. You can prove this by a simple compactness argument using the elementary diagram of $M$, augmenting the theory with a new constant symbol $\dot\theta$ and the assertions that the right theory holds. This theory is finitely consistent by an application of the Reflection theorem.
Second, some models $M$ of set theory cannot be realized directly as the $V_\theta$ of a taller model. For example, this is true in any pointwise definable model (a model in which every element is definable without parameters), since the larger model would then see that every element of $V_\theta$ is definable in $V_\theta$, which would contradict that the larger model thinks it is uncountable.
Third, every countable computably saturated model $M$ of set theory is realized as the $V_\theta$ of a taller model. Indeed, every countable computably saturated model of ZFC is isomorphic to one of its own $V_\theta$'s, and by looking at things from the perspective of that copy, the result obtains. Harvey Friedman was the first to prove results of this kind, concentrating at first on nonstandard models of PA. You can find an account of the argument in my recent paper with Victoria Gitman in "A natural model of the multiverse axioms," Notre Dame Journal of Formal logic, vol 51, 2010. One of the things we prove is that if $M$ is a countable computably saturated model of ZFC, then $M$ is also isomorphic to an element of $M$ that thinks it is an $\omega$-nonstandard model of set theory. Thus, every countable computably saturated model of ZFC exists as a nonstandard model inside another better model.
Fourth, it is possible to show that every countable transitive model of set theory can be end-extended to a (possibly nonstandard) model of V=L, which is well-founded to any desired countable height. The reason is that this assertion is true in $L$ itself, and the statement has complexity $\Pi^1_2$ in a code for the structure, so it is true in $V$ by Shoenfield absolutenss.