I think this does only work for so-called homological vector fields, i.e. vector fields of degree 1 which self-commute. Then you have $[v,v]=0$, which is the Maurer-Cartan equation in $T_{poly}$, and which insures that the corresponding derivative $d$ squares to zero.
Dealing with a Maurer-Cartan element you can simply use it to twist Kontsevich's formality $L_\infty$-quasi-isomorphism. When the Maurer-Cartan element is a vector field we know the explicit form of the first Taylor component of the twisted $L_\infty$-morphism: it is not HKR, but involve Bernoulli numbers (see e.g. https://www.math.ethz.ch/u/calaqued/research/LecturesDufloETH.pdf).
Anyway, the $k$-th taylor component of the $v$-twisted $L_\infty$-morphism will be given by the series $$ \phi_v^{(k)}(u_1,\dots,u_k):=\sum_{l\geq0}\frac{t^{l+1}}{(l+1)!}\phi^{(k+l)}(u_1,\dots,u_k,\underbrace{v,\dots,v}_{l~times}) $$ To conclude one just have to observe that $\phi^{(k)}$ preserves the grading given by the arity minus $2$ (arity means the number of arguments for poly-vectors and poly-differential operators). Therefore each time $v$ appears in the formula it decreases this degree by $1$.
EDIT: I include a comment into the main answer about the shape of $\phi^{(1)}_v$ in specific cases. Let me denote coordinates by $u^k$, and write $\partial_k=\frac{\partial}{\partial u^k}$ and $v=\sum_iv^i\partial_i$. We consider a matrix-valued one-form $\Xi$ given by $$ \Xi_i^j=\sum_k\partial_k\partial_iv^jdu^k, $$ and define $$ \Theta=\sum_{n>0}c_n\iota_{tr(\Xi^n)}. $$ Here $c_n$ are rational coefficients that do not matter.
Then $\phi_v^{(1)}$ is given by the precompsition of HKR with $e^{\Theta}$.
Now if I split the coordinates into even $x^i$ and odd $e^i$, and if I assume that $v=\sum_iv^i(x^1,\dots,x^m)\frac{\partial}{\partial e^i}$, then the matrix $\Xi$ has non-zero entries only in the right-up block. In particular it is upper triangular, the trace of any power of it is therefore zero, and thus $\phi^{(1)}_v=HKR$.