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 finite union of intervals $F\in A$, 
let the connected components of $F$ be $[a_0,b_0),\ldots,[a_k,b_k)$
with $p$ elements of $\{a_i,b_i:i\leq k\}$ added and $q$ removed.
Partition $F$ into its atomic subsets $H_0,\ldots,H_n$.
Choose a positive $\mu_A(H_i)\in R$ for each $i$, such that
$\sum_{i\leq n}\mu_A(H_i)=(p-q)\delta+\sum_{i\leq k}(b_i-a_i)$.
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.