Is there a group scheme G such that there is a proper class of non-isomorphic fpqc G-bundles over some fixed base? Famously, the fpqc topology is so big that no one set of fpqc covers of an affine scheme is cofinal in all fpqc covers. The Stacks Project gives a construction using field extensions with arbitrarily large infinite transcendence basis, which are all fpcq maps.
This means that it is difficult to see whether the collection of isomorphism classes of fpqc-locally trivial principal $G$-bundles, for a fixed group scheme $G$, on a given scheme form a set, rather than a proper class. If $G$ is nice (e.g. finite type/smooth/insert your favourite condition here), then standard reductions show that any fpqc bundle is in fact locally trivial in a far smaller topology, such as the étale topology. I think (from reading this thesis) that even if we are working with a group scheme over a field $k$ such that $G\to Spec(k)$ is fppf, then we should probably only get set-many isomorphism classes, as the stack of $G$-bundles is then algebraic. So any group scheme that would give a proper class of non-isomorphic bundles would have to be rather nasty. (Also, I believe we need to be in positive characteristic, to get maximum nastiness...)
My vague plan for constructing an example would be to consider fpqc $G$-bundles on $Spec(k)$ ($G$ a group scheme over $k$) via cocycles over the covers that the Stacks project describes at the above link. If a purely transcendental extension $k\to k\{I\}$ (where $I$ is a set of arbitrary cardinality) made $Spec(k\{I\}) \to Spec(k)$ a torsor (so the extension is 'Galois' for a suitable interpretation of that term), then one could consider homomorphisms $\alpha\colon Gal(k\{I\}) \to G$ and the associated $G$-bundles. It seems possible that one could figure out when two such bundles are isomorphic, possibly using a cardinality bound coming from data attached to $G$. Or not, I really don't know.
Can we make a call in either direction? Does the above sketch give proper class-many non-isomorphic bundles, or can we prove that for every fpqc group scheme $G$ and base scheme $X$ there is only a set of isomorphism classes of fpqc principal $G$-bundles on $X$?
 A: In more down-to-earth terms, you are asking whether there is always a set of $G$-torsors for the fpqc topology over a scheme $S$ such any every $G$-torsor for the fpqc topology over $S$ is isomorphic to one of those in the set. The answer is "yes" when $G$ is flat over $S$ (as is automatic over a field). 
The question asks about something being a "proper class", which I am told in the comment by Q. Yuan below means "class that is not a set", so the answer to the question in the title is then "no" under that flatness hypothesis on $G$. I don't see a reason for interest in torsors for for the fpqc topology on $S$ relative to a group scheme that isn't flat over the base, so I will regard the $S$-flatness of $G$ as a basic hypothesis one should impose.
The real content is in the following simple lemma:
Lemma: Let $A$ be a ring, $B$ an $A$-algebra, and $A'$ a faithfully flat $A$-algebra.  For an infinite set $I$, if the $A'$-algebra $B' = A' \otimes_A B$ admits a set of at most $|I|$ generators $($equivalently, $B'$ is a quotient of a polynomial algebra $A'[X_i]_{i \in I}$ over $A'$$)$ then so does $B$ over $A$.
This lemma and its proof are motivated by the more well-known fact (and its proof!) that if $B'$ is finitely generated (resp. finitely presented) over $A'$ then the same holds for $B$ over $A$ (EGA IV$_2$, 2.7.1.1), but the proof is easier for this controlled "infinitely generated" property.
Proof: Pick a set of generators of $B'$ over $A'$ indexed by a subset $J \subset I$. Write each as a finite sum of elementary tensors, and consider the set $S$ of elements of $B$ that arise in those tensors.  Since $I$ is infinite, it is clear that $|S| \le |I|$. Hence, it suffices to show that $B$ coincides with its $A$-subalgebra $B_0$ generated by the elements in $S$.  
By flatness of $A'$ over $A$, the natural map $A' \otimes_A B_0 \rightarrow A' \otimes_A B$ is injective.  But by design of $B_0$ in terms of $S$, this latter injection is also surjective.  Hence, by faithful flatness the inclusion $B_0 \hookrightarrow B$ is an equality too.
QED Lemma
Now we consider $G$-torsors for the fpqc topology on $S$, where $G$ is $S$-flat (hence faithfully flat, due to the identity section, but whatever).  It suffices to show that for each open affine $U = {\rm{Spec}}(A) \subset S$ (or really just the members of whatever fixed affine open cover you wish) there is a set's worth of isomorphism classes of $G$-torsors for the fpqc topology over $U$. That is, we can assume $S$ is affine, say with coordinate ring $A$.  By choosing finitely many affine opens in $G$ which cover the quasi-compact identity section, we can shrink $S$ a bit more so that there exists an open affine $V \subset G$ containing the identity section (so $V \rightarrow S$ is faithfully flat between affines). Let $I$ be an infinite cardinal so that each affine open subscheme of $G$ (e.g., $V$) admits a set of at most $|I|$ generators for its coordinate ring as an $A$-algebra (such an $I$ clearly exists); for what follows it would be enough to pick $I$ that "works" for the members of a single affine open covering of $G$. 
For any given faithfully flat $A$-algebra $R$, there is a set's worth of isomorphism classes of $G$-torsors for the fpqc topology over $S$ which admit an $R$-point (since by fpqc descent these are controlled by the descent data encoded in ${\rm{H}}^1(R/A, G)$; note that not all such descent may be effective when $G$ isn't relatively affine). 
Hence, it suffices to find a set's worth of such $R$ so that every $G$-torsor for the fpqc topology on $S$ splits over one of those $A$-algebras $R$ (i.e. admits an $R$-point over $A$). We will show that the set (!)  of isomorphism classes of faithfully flat $A$-algebras $R$ that admit a set of $A$-algebra generators of size at most $|I|$ does the job.
Let $E$ be a $G$-torsor for the fpqc topology over $S$. Thus, there exists a faithfully flat $A$-algebra $A'$ then $E(A')$ is non-empty, so $E_{A'} \simeq G_{A'}$ as $A'$-schemes.  In particular, $V_{A'}$ is an affine open subscheme of $E_{A'}$ admitting at most $|I|$ generators as an $A'$-algebra.  Its image in $E$ is quasi-compact, so is contained in a quasi-compact open $\Omega \subset E$.  The map $\Omega \rightarrow S = {\rm{Spec}}(A)$ is faithfully flat because after fpqc base change to $A'$ it becomes $\Omega_{A'} \rightarrow {\rm{Spec}}(A')$ that is faithfully flat (as $\Omega_{A'}$ is identified with an open subscheme of $G_{A'}$ containing $V_{A'}$).  
By quasi-compactness, $\Omega$ is covered by a finite set of affine open subschemes ${\rm{Spec}}(B_j)$ with $1 \le j \le n$.  Thus, $B = \prod_{j=1}^n B_j$ is a flat $A$-algebra that is faithfully flat since ${\rm{Spec}}(B) = \coprod {\rm{Spec}}(B_j) \rightarrow \Omega$ is surjective.  Hence, $E(B)$ is non-empty with $B$ faithfully flat over $A$.  We claim that $B$ admits a set of at most $|I|$ generators as an $A$-algebra (which will do the job).  It suffices to do the same for each $B_j$ separately.  
Now finally we do something that isn't formal nonsense, namely we invoke the Lemma: it suffices to show that $B_j \otimes_A A'$ admits a set of at most $|I|$ generators as an $A'$-algebra. But the spectrum of this $A'$-algebra is an affine open subscheme of $E_{A'} = G_{A'}$, so it suffices to show that every affine open $W$ in $G_{A'}$ admits a set of at most $|I|$ generators for its coordinate ring as an $A'$-algebra. By design, $G$ is covered by affine opens each of whose coordinate rings admits a set of at most $|I|$ generators as an $A$-algebra.  Thus, the same holds for $G_{A'}$ relative to $A'$ for the base-change covering, and consequently for any basic affine open of one of those. 
Recall the fact that for any two affine opens in a scheme, their overlap is covered by affine opens which are simulatenously affine open in each. Hence, the affine $W$ is covered by basic affine opens (relative to $W$!) whose coordinate rings each admit a set of at most $|I|$ generators as an $A'$-algebra. By quasi-compactness, $W$ is covered by finitely many of those. Since $I$ is infinite, the numerators in such fractional expressions give a set of size at most $|I|$ that generate the coordinate ring of $W$ as an $A'$-algebra.
