No.

Let's work with an Artinian local ring $R$. Let $X$ be a module satisfying $\mathrm{Ext}_R^i(X,R)\neq 0$ for all $i$. (These are plentiful; for example, assume $R$ is non-Gorenstein and let $X$ be the residue field.) Define a module $Q$ by "Serre's trick": take generators $\chi_1, \dots, \chi_r$ for $\mathrm{Ext}_R^1(X,R)$ and consider the short exact sequence corresponding to $(\chi_1, \dots, \chi_r) \in \mathrm{Ext}_R^1(X,R^r)$:
$$0 \to R^r \to Q \to X \to 0$$
Then $\mathrm{Ext}_R^1(Q,R) =0$, since the long exact sequence of $\mathrm{Ext}$ looks like
$$\cdots \to \mathrm{Hom}_R(R^r,R) \to \mathrm{Ext}_R^1(X,R) \to \mathrm{Ext}_R^1(Q,R) \to \mathrm{Ext}_R^1(R^r,R)=0$$
and the map $\mathrm{Hom}_R(R^r,R) \to \mathrm{Ext}_R^1(X,R)$ is cooked up precisely to be surjective. On the other hand, $\mathrm{Ext}_R^i(Q,R) = \mathrm{Ext}_R^{i}(X,R) \neq 0$ for all $i\geq 2$.

One can jazz this up a bit, using a result of Jorgensen–Şega (journal link): There exists a local Artinian ring $R$ and a family $\{M_s\}_{s\geq 1}$ of reflexive $R$-modules such that (among other things) $\mathrm{Ext}_R^i(M_s,R)\neq 0$ if and only if $1 \leq i \leq s-1$. They give a completely explicit construction of $R$ and the modules $M_s$. Taking one of the $M_s$ in place of $X$ above, one obtains modules $Q_{a,b}$ for which $\mathrm{Ext}_R^i(Q_{a,b},R)$ vanishes up to $i=a-1$, is nonzero for $a \leq i \leq b$, and vanishes again for $i \geq b+1$. Then taking direct sums gives essentially arbitrary behavior of vanishing and non-vanishing.

2more comments