No, I don't think so.
Lemma: Given $R$, $I$, $R_0 = R/I$, $A$, $A_0 = A/I$. Assume $2$ is invertible in $R_0$. If the answer to the question is "yes" then the image of the map $Pic(A) \to Pic(A_0)$ contains all $2$-torsion of $Pic(A_0)$.
Proof: Namely, suppose $L_0$ is an invertible $A_0$-module of order $2$ in the Picard group of $A_0$. Choosing an isomorphism $\psi : L_0 \otimes_{A_0} L_0 \to A_0$ we can set $B_0 = A_0 \oplus L_0$ with multiplication defined by $\psi$. Then $A_0 \to B_0$ is finite etale because $2$ is invertible in $A_0$ by the branched covering trick, whence $B_0$ is smooth. If we can find $B$ as desired, then we see that $\det_A(B) \in Pic(A)$ lifts $L_0$. QED
So now we just have to find $R$ and $A$ where the $2$-torsion in $Pic(A_0)$ does not lift. Read on if you want to see what I came up with. To understand it, you might have to recall some facts about the deformation to the normal cone. See example at the end for an explicit case.
Say $R$ is a complete discrete valuation ring with uniformizer $t$ and residue field $k$. Let $n \geq 2$ and let $C \subset \mathbf{A}^n_k$ be an irreducible closed smooth subscheme such that $Pic(C)$ has lots of $2$-torsion. Denote
$$
b : T \longrightarrow \mathbf{A}^n_R
$$
the blowing up of $C$ viewed as a closed subscheme by identifying $\mathbf{A}^n_k$ with a closed subscheme of $\mathbf{A}^n_R$. Then we have
$$
V(t) = E + X
$$
where $E$ is the exceptional fibre of $b$ and $X$ is the strict transform of $\mathbf{A}^n_k$, i.e., $X$ is the blowing up of $C$ in $\mathbf{A}^n_k$. Since $E$ is an anti-ample divisor, we see that $X$ is an ample divisor on $T$. Hence we see that
$$
U = T \setminus X = \text{Spec}(A)
$$
is an affine scheme (complement of an ample divisor on a scheme projective over an affine scheme is affine). It is smooth over $R$ as the non-smooth locus of $b$ is $E \cap X$. The special fibre $U_0 = \text{Spec}(A_0)$ is an $\mathbf{A}^c$-bundle over $C$ where $c$ is the codimension of $C$ in $\mathbf{A}^n_k$. Thus we see that $Pic(A_0) = Pic(U_0) = Pic(C)$ has nonzero $2$-torsion elements.
Finally, the generic fibre of $\text{Spec}(A)$ is $\mathbf{A}^n_K$ where $K$ is the fraction field of $R$. Since $t$ is a prime element of $A$ this implies that $\text{Pic}(A)$ is trivial.
Example: If $C$ is a hypersurface $C = V(\overline{f})$, then I think you just get $A = R[x_1, \ldots, x_n, y]/(ty - f)$ where $f$ is a lift of $\overline{f}$. You can verify all the properties of this directly, I guess.
