Despite Will Sawin's answer in comment, it was not so obvious for me at first glance, so I give a detailed (at least for $X$ affine) proof here.

**Lemma.** Let $A$ be a $K$-algebra and let $A\to B$ be fpqc. Then the following sequence is exact $$
A((t))\to B((t))\rightrightarrows (B\otimes_{A}B)((t)).
$$

**Proof.**
Since $A\to B$ is fpqc, Grothendieck's fpqc-descent shows that the sequence
$$
A\to B\rightrightarrows B\otimes_A B
$$
is exact. Consider the sequence
$$
A[[t]]\to B[[t]]\rightrightarrows (B\otimes_{A}B)[[t]].
$$
We claim that this sequence is also exact.

The first arrow $A[[t]]\to B[[t]]$ is injective since $A\to B$ is injective and since $A[[t]] = \prod_{n\in \mathbf{N}}A$
and $B[[t]] = \prod_{n\in \mathbf{N}}B$ as $A$-modules.
It remains to show that for $f(t)\in B[[t]]$,
$$f(t)\otimes 1 = 1\otimes f(t)\in (B\otimes_{A}B)[[t]] \quad\Longrightarrow \quad f(t)\in A[[t]].
$$
Suppose that $f(t) = \sum_{n\in \mathbf{N}} b_n t^n\in B[[t]]$ satisfies the condition $f(t)\otimes 1 = 1\otimes f(t)$. It is enough to show that $f(t) \mod{t^N} \in A[t]/t^N$ for each $N\ge 1$. Indeed, if we tensorise the first exact sequence by $\otimes_{A}A[t]/t^N$, we get
$$
A[t]/t^N\to B[t]/t^N\rightrightarrows (B\otimes_{A} B)[t]/t^N,
$$
which implies that $b_n\in A$ for $n < N$. Hence $f(t)\in A[[t]]$.

Inverting $t$, we deduce the exactness of
$$
A((t))\to B((t))\rightrightarrows (B\otimes_{A}B)((t)).
$$
**End of Proof**

Now if $X = \mathrm{Spec}\, C$ is an affine $K$-scheme and A,B are as above, then the exactness of the following sequence follows from the lemma
$$
\mathrm{Hom}(C, A((t)))\to \mathrm{Hom}(C, B((t)))\rightrightarrows \mathrm{Hom}(C, (B\otimes_{A} B)((t))).
$$
Hence, if $X$ is affine, $LX$ is automatically an fpqc sheaf. The general case follows from the standard procedure of glueing.