Yes. For every suitable theory (powerful enough to handle basic arithmetic) there is a statement $G$ in the language of that theory such that one can proof that no proof of $G$ exists. Moreover, $G$ is true, hence a proof of $\neg G$ should not exist either. However, one is not absolutely sure of the latter because one cannot prove satisfactorily that the theory is sound.
Look for Gödel and incompleteness theorem.
Actually, this is possibly not exactly what you talk about later in the dialog. The $G$ mentioned above is provably not provable and it is (assuming soundness of arithmetic) provably not refutable. Hence it is not possible to show that $G$ is provable and it is not possible to sow that $G$ is refutable (because it is impossible to show wrong things). But if the Asker asks: "Can we show $G$? Can we refute it?", the Answerer might reply "We know that we can'T" and not "It is impossible to know that". Do you really want this extra level of uncertainty or was that just a bad formulation?
In that case: Loosely speaking, $G$ is obtained by formalising "$x$ is a proof of $y$" and introducing Gödel numberings to make provabilty a question about arithmetic properties. Then comes diagonalization and one has
- a map $\Gamma$ from the set of formulas to $\mathbb N$
- a represented proof predicate $P(a,b)=\Gamma^{-1}(a)\mathrm{\ is\ a\ proof\ of\ }\Gamma^{-1}(b)$
- thus a represented theorem predicate $T(b)=\exists a\colon P(a,b)$
- a represented diagonalization $D(a,b,n)=\Gamma^{-1}(a)\mathrm{\ is\ }\Gamma^{-1}(b)\mathrm{\ with\ all\ occurences\ of\ }x\mathrm{\ replaced\ by\ }\underbrace{ S\cdots S}_n0$
- finally a represented predicate $G_0(n)=\exists b\colon D(n,b,n)\land T(b)$ The last step is to consider $n:=\Gamma(\neg G_0(x))$ and the statement $G:=G_0(n)$. With all reinterpretations, if $G$ is true, then $\neg G_0(n)$ is provable - contradiction. Hence $G$ is false and $\neg G=\neg G_0(n)$ is true but not provable.
Now replace $T$ ("provable") above with "decideable": $T'(a)=T(a)\lor T(\Gamma(\neg\Gamma^{-1}(a))).$ Does that help?