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 $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$
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 $$ A[[t]]\to B[[t]]\rightrightarrows B[[t]]\otimes_{A[[t]]}B[[t]]. $$ 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 $f(t)\in B[[t]]$, $$f(t)\otimes 1 = 1\otimes f(t)\in B[[t]]\otimes_{A[[t]]}B[[t]] \quad\Longrightarrow \quad f(t)\in A[[t]]. $$ 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 $$ A[t]/t^N\to B[t]/t^N\rightrightarrows B[t]/t^N\otimes_{A[t]/t^N} B[t]/t^N, $$ which implies that $b_n\in A$ for $n < N$. Hence $f(t)\in A[[t]]$.
Inverting $t$, we deduce the exactness of $$ A((t))\to B((t))\rightrightarrows B((t))\otimes_{A((t))}B((t)). $$ 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 is obviously exact $$ \mathrm{Hom}(C, A((t)))\to \mathrm{Hom}(C, B((t)))\rightrightarrows \mathrm{Hom}(C, B((t))\otimes_{A((t))} B((t))). $$ Hence, if $X$ is affine, $LX$ is automatically an fpqc sheaf. The general case follows from the standard procedure of glueing.