I am interested in calculating $H(X|Z)$ for $Z=X+Y$ is a noisy version of $X$.
Initially, I was thinking since $Z$ is just a noisy version of $X$ then $H(X|X+Y) = H(Y)$ is just the uncertainty in the additive noise applied to $X$.
But, when I played through an example, I found different results.
I am considering
$$P(X) = \begin{cases} \frac{1}{2}, &X=0\\ \frac{1}{2}, &X=1 \end{cases}$$
$$P(Y) = \begin{cases} (1-p), &Y=0\\ p, &Y=1 \end{cases}$$
Then
\begin{array}{llll|ll} X &Y &Z &prob \\ \hline 0 &0 &0 &\frac{1}{2}(1-p)\\ 0 &1 &1 &\frac{1}{2}p\\ 1 &0 &1 &\frac{1}{2}(1-p)\\ 1 & 1 &2 &\frac{1}{2}p\\ \end{array}
From this we can find $P(Z)$,
$$P(Z) = \begin{cases} \frac{1}{2}(1-p), &Z=0\\ \frac{1}{2}, &Z=1\\ \frac{1}{2}p, &Z=2 \end{cases}$$
Then calculating $P(X|Z)$,
\begin{array}{l|ll} P(X|Z) &0 &1 \\ \hline 0 &1 &0 \\ 1 &p &(1-p) \\ 2 &0 &1 \end{array}
Now Finding $H(X|Z)$,
$$H(X|Z) = \sum_z p(z) H(X|Z=z)$$
$$= p(z=1)H_b(p,1-p)$$
$$= \frac{1}{2}H(Y)$$
Is this the correct calculation? I thought that I had read somewhere that $H(X|X+Y) = H(Y)$, but according to this example its not exactly true.
It's not true that $H(X|X+Y) = H(Y)$ - your calculation is a counterexample. What is true is that if $X$ and $Y$ are independent, then $H(X+Y|X) = H(Y)$ (exercise - show this), perhaps you're getting confused with this.
In fact, we can show that if $Y$ and $X+Y$ are not independnet, then $H(X|X+Y) < H(Y).$
Since $(x,y) \mapsto (x, x+y) $ is a bijection, $H(X,Y) = H(X,X+Y)$, and by the chain rule, $$ H(X,Y) = H(X, X+Y) = H(X+Y) + H(X|X+Y).$$ But, by the chain rule, it's also true that $H(X,Y) = H(X|Y)+ H(Y).$ But then $$ H(X|X+Y) = H(Y) + H(X|Y) - H(X+Y)\\ = H(Y) + H(X+Y|Y) - H(X+Y) \\ = H(Y) - I(Y;X+Y).$$ Due to non-independence the final mutual information is non-zero, showing that $H(X|X+Y) < H(Y)$.