Yes, by compactness. 

Let $R$ denote your favorite hyperreal ordered field 
and let $\delta\in R$ be a positive infinitesimal. 
Let $\mathcal{E}$ denote the set of all 
(standard) finite Boolean subalgebras of $\mathcal{P}([0,1))$.
For every $A\in\mathcal{E}$,
let $\lambda_A(I)$ be the (exact) length of $I$ 
for all half-open intervals $I\in A$; 
for all open or closed intervals $I\in A$, 
respectively subtract or add $\delta$ to the length of $I$ 
to define $\lambda_A(I)$;
let $\lambda_A(S)=\delta$ for all singletons $S\in A$.
Extend $\lambda_A$ to a probability measure $\mu_A$ on $A$.

(Specifically, for each minimal interval $I\in A$
that is not an atom of $A$, 
partition $I$ into its atomic subsets $H_0,\ldots,H_n$
such the singletons among $H_0,\ldots,H_n$ are $H_0,\ldots H_{m-1}$.
Choose a positive $\mu_A(H_i)\in R_1$ for each $i\geq m$, such that
$\sum_{i=m}^n\mu_A(H_i)=\lambda_A(I)-m\delta$. For all atomic
$K\in A\cap\mathrm{dom}(\lambda_A)$, define $\mu_A(K)=\lambda_A(K)$.
Now extend $\mu_A$ from the atoms to all of $A$.)

Let $U$ be a fine ultrafilter on $\mathcal{E}$.
("Fine" means that $\{B\in\mathcal{E}:A\subseteq B\}\in U$
for all $A\in\mathcal{E}$). 
The ultraproduct measure $\mu_U$ is $R^U$-valued and has 
the two properties you seek.