The answer is no, because such ultrapowers are always $\aleph_1$-saturated, but $\mathbb{R}$ is not.

More concretely, the ultraproduct will be an ordered field with uncountable cofinality — every countable set is bounded above. To see this, observe first that because the order $x<y$ is definable in the real field, if this field were realized as an ultraproduct, then almost every $K_i$ will have to be an ordered field. It now follows that every countable sequence in the ultraproduct will be bounded, because like Hausdorff we can climb above any countable sequence of functions. Specifically, for any countably many $f_0,f_1,f_2,\ldots\in\prod_i K_i$, let $f(i)=\max_{k\leq i} f_k(i)+1$ in $K_i$, a finite supremum on each coordinate. The function $f$ is eventually above every $f_k$ on a tail, and so $[f]$ is above every $[f_k]$ in the ultraproduct.

But not every countable set in $\mathbb{R}$ is bounded, since for instance $\mathbb{Z}$ is unbounded in $\mathbb{R}$.

Incidentally, you asked about infinitesimals, and these exist in the ultraproduct by the same kind of reasoning. Just let $e(i)=1/i$ in $K_i$, and it follows that for any specific natural number $n$ we have $e(i)<1/n$ for almost all $i$, and so $[e]$ is infinitesimal in the ultraproduct. In fact, the positive elements of the ultraproduct have uncountable coinitiality, meaning that every countable set of positive elements is bounded below. Given $f_0,f_1,\ldots$ of positive elements, let $e(i)$ be the minimum of $f_k(i)$ for $k\leq i$, and then $0<[e]\leq[f_k]$ in the ultraproduct for every $k$.

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