$\newcommand{\B}{\mathcal B}$
First here, $[0,1]^d$ is a Polish space (i.e., a separable complete metric space). So, $[0,1]^d$ is a [Radon space][1], and hence any (Borel) probability measure is Radon. So, you did not have to say that the probability measure $\pi$ on $[0,1]^d$ is Radon. (Also, you did not have to assume that $\pi$ is singular.) 

Anyway, [it follows that][2] for each $i=1,\dots, d$ your probability measure $\pi$ admits a regular conditional probability distribution $[0,1]\times\B_d\ni(t,A)\mapsto\nu_i(t,A)\in[0,1]$, such that the map $[0,1]\ni t\mapsto\nu_i(t,A)\in[0,1]$ is $\B_1$-measurable for each $A\in\B_d$, the map $\B_d\ni A\mapsto\nu_i(t,A)\in[0,1]$ is a probability measure for each $t\in[0,1]$, and 
$$\pi(A\cap p_i^{-1}(B))=\int_B\nu_i(t,A)\pi(p_i^{-1}(dt))
$$
for all $A\in\B_d$ and $B\in\B_1$, where $\B_k$ is the Borel $\sigma$-algebra over $[0,1]^k$ and $[0,1]^d\ni(x_1,\dots,x_d)\mapsto p_i(x):=x_i\in[0,1]$. 

So, you can let $\mathbf P_\pi [\cdot \mid x_i=\alpha]:=\nu_i(\alpha,\cdot)$ for all $\alpha\in[0,1]$. 

  [1]: https://en.m.wikipedia.org/wiki/Polish_space#Radon_spaces
  [2]: https://en.m.wikipedia.org/wiki/Regular_conditional_probability#Definition