I'm considering a problem around the moduli of perfect complexes.
Let's consider a $X$ smooth proper over Spec($R$), $R$ is a DVR, $char=0$, equipped with a perfect complex $\mathcal F$ on $X_{K(R)}$.
Is there any obstruction that prevents the existence of a perfect complex $\mathcal F'$ on X such that $\mathcal F'|_{X_{K(R)}}=\mathcal F$？
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Thomason and (the ghost of) Trobaugh proved the following remarkable theorem.
Theorem (Proposition 5.2.2 in [TT]). Let $X$ be a quasi-compact and quasi-separated scheme, let $U\subseteq X$ be a quasi-compact open subscheme, and let $j\colon U\hookrightarrow X$ be the inclusion. Let $F^\bullet$ be a perfect complex on $U$. Then $F^\bullet$ is isomorphic in the derived category $D(\mathcal{O}_U\text{-mod})$ to the restriction $j^* E^\bullet$ of some perfect complex $E^\bullet$ on $X$ if and only if the class $[F^\bullet]$ in $K_0(U)$ is in the image of the map $$ j^* \colon K_0(X) \to K_0(U). $$
Now if $X$ is regular (as in your situation) then $K_0(X) = K_0(\mathrm{Coh}(X))$. Since coherent sheaves always extend (see e.g. Exercise II 5.15, p. 126 in Hartshorne), the map $j^*\colon K_0(X)\to K_0(U)$ is surjective. So the answer to your question is that a perfect complex on $U = X_{K(R)}$ always extends to a perfect complex on $X$.
Edit. I can't help but add that the final paragraph of the introduction to [TT] (very relevant to the discussion) is one of the most moving passages in all of mathematics.
[TT] Thomason, R. W.; Trobaugh, Thomas. Higher algebraic $K$-theory of schemes and of derived categories. The Grothendieck Festschrift, Vol. III, 247--435, Progr. Math., 88, Birkhäuser Boston, Boston, MA, 1990. MR1106918
